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.:
Here's an actual example of measurements/calcs:
0.29A rms 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.
Sounds like you have a lot of time on your hands.
Here is a link to some useful charts:
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
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
:I cant answer you but a meter that is fun and usefull is a
: apx 25$, it lets you plug in an apliance and measures very
: time, watt, amp, power factor, Kwh used and the hours, V. Hz
: you will easily be able to audit usage of anything 120v you
plug into it
: and find the hogs to show the hogs. It is also very good on low
: equipment on standby, such as the tv off. It measures over a
: period. You would be suprised how older things can cost 1$ a
: 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.
: made me get a new frige that uses 1/6th the power. They have
: independantly tested very accurate, somethimes Radio shack has
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.
: 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:
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
: The power company puts power factor correction equipment on
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.
for those of you who want to "geek out" on power factor correction &
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)
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)
: BobK207 wrote:
: > for those of you who want to "geek out" on power factor
: > reactive power,
: > here are couple of links that give understandable
: > http://home.earthlink.net/~jimlux/hv/pfc.htm
: > http://www.nepsi.com/powerfactor.htm
: > the "best" power factor correction is achieved by adding
: > capacitors at each inductive load (motor)
: > cheers
: > Bob
: Nothing new if one paid attention in his/her HS physics class.
: In real world MOST electrical load is inductive which makes
: current by certain amount. Reactive power is false power(wasted
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
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
If anyone has an understanding of this, I'd love to hear it... been
wondering about this one for a while. :)
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
: no useful work in your electric motor, it's doing work
: I'm sure it's an obvious point but the answer isn't evident to
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
: 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
: 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...
: Heat loss of the electrons rushing into and out of the
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
Capacitor stores electrons.
Inductor creates current flow from a collapsing field, resists
them during the building of hte field. Limited by the resistance
: If anyone has an understanding of this, I'd love to hear it...
: wondering about this one for a while. :)
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
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.
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.
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.
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
: >... If the power company is feeding you 1000W on a straight
: >and your motor is seeing just 700W of useful work on a PF*V*A
: >there are 300W of energy that have "disappeared". I guess the
: >is so simple that the answer is obvious: The energy is
: >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
: With a 0.7 PF, 5.83/0.7 = 8.33 amps flow, so the wiring would
: 8.33^2x0.1 = 6.9 watts, and you pay for 706.9 watts. Your only
: is the difference, 6.9-3.4 = 3.5 watts. The power company is
: because they lose more power in their wiring, but they usually
: complain, in the case of houses.
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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