Power Factor & kWH?

[chocolatemalt]

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.

[Pop]

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

[Richard J Kinch]

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

[BobK207]

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

here are couple of links that give understandable explanations

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

cheers Bob

[Tony Hwang]

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

[Pop]

"chocolatemalt" wrote in message news: [e-mail]... : 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 ? 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!

[Pop]

: > here are couple of links that give understandable explanations : >

: >

> > >

: > 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.

[chocolatemalt]

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. :)

[nicksanspam]

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

[keith]

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]

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).

[keith]

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.

[Pop]

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. :)

[chocolatemalt]

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

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.

[nicksanspam]

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

[chocolatemalt]

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&#39;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.

[larry]

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&#39;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

[Pop]

"keith" wrote in message news: [e-mail]... : On Sat, 12 Nov 2005 19:54:52 -0500, Pop wrote: : : >

: > "chocolatemalt" wrote in : > message : > news: [e-mail]... : > : 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&#39;d : > be surprised how far off that same 100W calc is if the load is : > inductive. Depending, I&#39;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&#39;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&#39;m sure the UPS ain&#39;t all that : > accurate as a rule either. How&#39;s that for a scientific calibration : > check ? : > Also, if you&#39;re playing with duty cycle, you don&#39;t multipy by : > 24 x 7 etc.; that&#39;s a 100% duty cycle on your assumption of everything : > having a power factor of 1.00. : > For an electric bulb though, you&#39;d be real close. But : > refrigerator, furnace, flourescent, things like that it&#39;s quite a : > different story. : > It&#39;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&#39;t recall what I meant to say but if I could do that, I&#39;d go into business selling the idea to others!! Typo obviously . Err, not, it wasn&#39;t either! Send me $11.95 in a SASE and I&#39;ll tell y&#39;all how ta do it! ;-} Now, where&#39;d I lay that gizmo? Hmmm ...

Regards,

Pop

[Pop]

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

: > A little trig is sufficient, IMO. : : Simple algebra is enough, depending on what you&#39;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.

[Pop]

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&#39;t : complain, in the case of houses. : : Nick : Yeah, I"ve since come across some info that indicates a " very bad" PF would be in the order of single-digit percentages, as in a few % or so. I was surprised it could get that low, but I guess it can.

Besides, it&#39;s not jsut heat lost to the wires & system; it&#39;s the fact that the voltage and current never reach peaks at the same point in time and thus IE never represents true power without including a PF. Apparently most of the loss is in the energy required to build the flux fields, interference to it, and then offset by the collapsing field, which is creating voltage by trying to increase current, etc etc etc.. Everybody seems to be missing the phase relationship between the voltage and current thru a reactive load.