# Power Factor & kWH?

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• posted on November 12, 2005, 12:24 am
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
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<%-name%>
• posted on November 12, 2005, 12:58 am

Sounds like you have a lot of time on your hands.
Here is a link to some useful charts: http://www.mrelectrician.tv/conversioncharts/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.
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• posted on November 12, 2005, 2:02 am
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.
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• posted on November 12, 2005, 1:34 pm
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
: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. :
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• posted on November 13, 2005, 1:59 pm
On Sat, 12 Nov 2005 08:34:10 -0500, Pop wrote:

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

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• posted on November 12, 2005, 1:33 pm
.... : : 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/conversioncharts/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. :
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• posted on November 12, 2005, 7:35 pm

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.
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• posted on November 12, 2005, 8:14 pm
chocolatemalt writes:

More conveniently, 1 watt-year costs 1 US dollar. But those days are passing.
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• posted on November 12, 2005, 10:44 pm
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/power_factor_correction_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
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• posted on November 12, 2005, 11:45 pm
BobK207 wrote:

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
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• posted on November 13, 2005, 1:00 am
: 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/power_factor_correction_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.
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• posted on November 13, 2005, 1:23 am

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. :)
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• posted on November 13, 2005, 1:05 pm

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
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• posted on November 13, 2005, 4:29 pm
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. :)
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• posted on November 13, 2005, 8:10 pm

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.
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• posted on November 13, 2005, 9:13 pm

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
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• posted on November 13, 2005, 11:05 pm
snipped-for-privacy@ece.villanova.edu wrote:

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.
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• posted on November 14, 2005, 4:44 am
chocolatemalt writes:

No, there is still an economic penalty to the current capability of the generation plant being used up for no delivered power. Capital is tied up without producing power for the customer.
This may be more costly than the line losses today.
It is not much different than the peak demand charges that even small commerical customers now endure, and which ought to be paid by everyone. The true cost of electric power is not just fuel and transimission, but the capital investment for peak capacity.
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• posted on November 14, 2005, 12:53 am
snipped-for-privacy@ece.villanova.edu wrote:

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
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• posted on November 14, 2005, 2:22 am
: : >... 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 : 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's not jsut heat lost to the wires & system; it'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.
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