how is power usage of CF's calculated?

how is power usage of CF\'s calculated?

Hi folks,

I"m holding 2 different GE CF bulbs in my hand at the moment. A spiral one that is labeled 120VAC, 230ma, 15W and a non-spiral one labeled 120VAC, 240ma, 14W. How can one be 230ma and 15 watt and the other 240ma and yet only 14w?

Also, is the ballast current counted in the power usage? Is that 14w through the tube or through the whole thing. It doesn't really way very much specifically, but the fact that at the same voltage one can draw more amps but fewer watts makes me suspicious /ubbthreads/images/graemlins/wink.gif

Re: how is power usage of CF\'s calculated?

Power factor?

*remembering my EE*

The 15W would have a power factor of 0.54
While the 14W would have a power factor of 0.49

Re: how is power usage of CF\'s calculated?

I can't answer your question, but I have yet to see a CF only pulling it's rated power. All the ones I have checked pull about 10% to 100% more power then rated.

So if I had to guess, I would say they are rated at the lamp, not including ballast.

Just taking your lamps for instance, 240mA at 120v is 28 watts. Yet the light is rated 15w? How can they call it a 15w light if it is pulling almost 30w? I have some 4w candelabra ones that pull about 8w and I have some other 5w candelabra base that pull 7w go figure.

Re: how is power usage of CF\'s calculated?

Hello there James,

You have to remember that if there is an angle between
current and voltage you dont just multiply the current
times the voltage to get the consumed power.
Assuming a sine wave current,

P=E*I*cos(angle)

If you solve for the angle knowing E and I, you
get approximately 60 degrees for the current angle.

The above is still a simplification of the formula to
follow,and above does assume pure sine waves however, but
it's entirely possible that these devices draw a pulsed
wave current, in the shape of a pulse but with a sine
wave envelope. The only way to calculate the power
in this case is on an instantaneous basis, with

Pavg=I{0 to T} of [e*i]/T dt

where
I{0 to T}
indicates the intergral of the function from t=0 to t=T,
where T is the period,
and
e is the instantaneous voltage (usually a function
of time t),
and
i is the instantaneous current (usually a function
of time t also).

To calculate this integral, you have to first know
both functions e and i, and the period T.
You can get this info by looking at scope waveforms
of both the current and the voltage.

So in the end, the power calculation isnt simply the
multiplication of E (voltage) times I (current), but
for the most general case is an integral. It is
however quite possible to estimate this integral using
a formula such as Simpsons rule or something like that,
but you still need scope waveforms to start with.

It's also possible that they didnt spec the correct
wattage for a given device, and it would be nice to
double check this using the above method.

Take care for now,
Al

Re: how is power usage of CF\'s calculated?

Ummm, I think I know what you are talking about. But when I meter them I am using a Fluke 87 true RMS. I know it reads much higher on my cheap meter, so I thought it took that in to consideration with how it was metering.

I think basically what you're saying is that it doesn't use the full 250mA the entire time it is on, thus reducing the overall power used, but max it is at 250mA?

Re: how is power usage of CF\'s calculated?

Hi there Brock,

I see you have your web site arranged so everything can be found
quickly in one place. That's very nice.

Your idea of the current not always being the same is
sort of correct because the two waves (voltage and current)
are sine waves, so they never stay at the same level, so the
instantaneous power delivered is never constant.

I hope this doesnt come off too 'authoritative', i just want
to give a reasonable answer to one of life's mysteries

The interesting thing is that the RMS voltage and current
measurements are considered to be constant. The catch is
that their values only have meaning when dealing with a
purely resistive load. This means that if we dont have
a resistive load, the meaning of the two cant be applied
to calculate power, unless of course we know two more things:
the angle between current and voltage, and they are both
sine waves.

This means that
if we know the RMS voltage and current through a measurement
with the meter we can calculate power simply by multiplying V * I
but only if the load is a pure resistance. That's the unfortunate
catch; if the load is anything other then a resistance we
still cant calculate the power knowing only the RMS current
and voltage.
The reason for this is that the RMS voltage value
(or peak, or average) is a property of the voltage alone,
and the RMS current value is a property of the current alone,
whereas
the average power (usually just called 'power') is a property
of both the voltage and the current taken simultaneously.
This is because power is not only dependent on the voltage
and current amplitudes, it's also dependent on the relationship
between the two. This in turn means in order to measure power
we would need a single instrument to make the simultaneous
measurements of voltage and current.

When we take the two instantaneous values simultaneously, we
multiply each point on the current waveform times each
corresponding point on the voltage waveform and generate
a third waveform. This third waveform represents the instantaneous
power. This really is simply current times voltage, but it's
done on a point by point basis, for every point in time.
Once we get this waveform, we then take it's average value.
The average value of this wave is the average power, commonly
called 'power'.
There is no way to reduce the calculation without knowing
each individual point in both waves for the most general case.


This might sound hard to do, and yes it's harder to do then
simply multiplying volts time amps, but it's not really as
hard as it sounds. If we pick 20 or so time points
equally spaced across a half cycle, we can multiply
the current points times the voltage points and add them
all up and divide the result by 20 and come up with a
decent accuracy for most purposes. It's also interesting
to do this on graph paper, drawing two sine waves
out of phase. If we draw the third resulting wave, it
looks more peaked then the other two waves because it's
not a sine wave.

If the two waves are sine waves, then we can apply the
simple formula:

P=I*E*cos(angle)

but we do have to know the angle.

To see how much error can occur without considering the angle,
lets see what would happen if we had an angle of 90 degrees...

If we had 120 volts rms and 1 amp rms, 60Hz lets say,
E*I equals 120. Is this 120 watts?

Since cos(90) equals zero,
E*I*0 equals 0.

Note that what looked at first like 120 watts turned out to
be in fact no power at all! This is close to what a real
inductor or capacitor would show.
Note also that the RMS voltage is still 120 volts, and the
RMS current is still 1 amp too, but the power in this case is 0.
To show how true this is, we could compare the temperature of
say a capacitor that draws 0.83 amps at 120vac to a 100 watt light
bulb. The 100 watt light bulb gets way too hot to touch while
the capactor doesnt even get warm yet they both draw the same
current


CONCLUSIONS

[1]
If we are dealing with pure dc or pure resistance, we can
multiply volts times amps to get the power.

[2]
If we are dealing with pure sine waves, we can use the
cosine of the angle between current and voltage to help
determine the true power, but we do have to have a way
of measuring the angle.

[3]
If we are dealing with an arbitrary waveform (triangle, etc)
we have to multiply the amplitudes of the
voltage wave by the amplitudes of the current wave for
each equally spaced point in time and
sum the results and divide by the number of points to
get an approximation of the true power.


One last thing to mention is that power meters have two inputs,
one for voltage and one for current, so that they can measure
both values at the same time and do the math.


Take care,
Al

Re: how is power usage of CF\'s calculated?

Hello again,

I've posted a method for approximating the power
in a circuit that has a reactive load, similar
to what the curly fluorescent lamps must be
drawing.
The topic is
"Compute real power in an ac reactive load".

Take care,
Al

Re: how is power usage of CF\'s calculated?

MrAl, it's a shame that we have to bother with all that RMS calculation, without that I could build you coils that tapped zero point energy and generated free power /ubbthreads/images/graemlins/grin.gif But with that it turns out all I'm generating is heat /ubbthreads/images/graemlins/icon15.gif

No, really, thats terrific and exactly what I wanted to know. I'll have to play with the math a bit but at least now I can understand how to do it.

Thanks again.

Re: how is power usage of CF\'s calculated?

I checked out philips specs on their CF bulbs - most have a PF in the 0.5-0.6 range. I wouldn't be suprised if what you are seeing is due to the PF of the bulbs (assuming measurement is correct and all).

One of my EE professors was saying that power companies had a hell of a time when switching power supplies started being made en masse (like for huge labs or large office building), since their power factor was so skewed compared to a linear supply (transformer/recitifier/regulator). The power companies have to balance their transmission source to the load, to get the PF as close to 1 as possible. A switching power supply is more capacitive, so skews the PF quite a bit (from what I remember, I may be wrong about this).

He also told us that you could "steal" power by running loads through a large capacitor (making the load more capacitive), since the power company only charges for "real" power used.

This is what I remember - but I may be forgetting something, or just remembering something wrong (feel free to correct me if so).

Re: how is power usage of CF\'s calculated?

[ QUOTE ]
evan9162 said:
He also told us that you could "steal" power by running loads through a large capacitor (making the load more capacitive), since the power company only charges for "real" power used.


[/ QUOTE ]

From The Art of Electronics by Horowitz and Hill:

"Power factor is a serious matter in large-scale electrical power distribution, because reactive currents don't result in useful power being delivered to the load, but cost the power company plenty in terms of I^2*R heating in the resistance of generators, transformers, and wiring. Although residential users are only billed for "real" power [Re(VI*)], the power company charges industrial users according to the power factor. This explains the capacitor yards that you see behind large factories, built to cancel the inductive reactance of industrial machinery (i.e., motors)."

Re: how is power usage of CF\'s calculated?

So does that mean I was right? (That's what I'm reading that snippet to mean)

Re: how is power usage of CF\'s calculated?

Hello again,

James S:
Sorry about all the math and the somewhat long procedure, i tried to find
the easiest way to find the true power using only a meter. It's just
not *that* easy, not as easy as i had hoped. I guess some things are
just that way; there's no way to make it simpler unless we dont mind
shelling out the bucks for an actual power meter that we could connect
and do a direct measurement without calculating anything. I dont think
they are that cheap though unless someone here might know of one?

evan9162:
From the current and voltage and the wattage rating, i was guessing something
around 0.5 too. I have several of these things around here i just didnt get
around to measuring any yet I hope to do that at some point just to verify
their specs.
What your EE prof must have been talking about is pf correction, which usually
starts with a serious pf problem and attemps to correct it. The correction
by adding reactive components usually means the out-of-phase current is
brought closer to zero, which reduces the current draw for a given load power.
Reducing the current draw in turn reduces the wire heating, which could
be alot, and so reduces the electric bill. Adding a capacitor just for
the heck of it wont really do anything except reduce the voltage getting
to the load (if that helps) which will reduce power consumed, but also
might not allow the load device to operate properly. The interesting thing
here is that the real power usually has to stay the same for a given device
to operate normally, so no matter what is added it doesnt help much unless
there is a serious problem with pf to begin with which causes wire heating
or other current related problems. As far as stealing power though, there
isnt any way to do that because the power meters the elect company uses
measures true power, and true power (energy) is what actually does anything
useful (like light a bulb, turn a motor against torque, etc). If you reduce
true power you reduce the usability of a device also. A good example
might be an electric heater...if you reduce the power entering the load
by connecting a cap in series, you do reduce the current draw by the load,
but you also reduce the heating effect which effectively makes the heater
look like a smaller less expensive heater

Current and voltage arent very much when taken alone. It's almost like
they dont really exist by themselves, because they dont do anything by
themselves. You can have 1000 amps flowing, but if there is no voltage
there is no real power.
In an ac circuit it is almost the same, except there can be high voltage
and high current (measured with a meter) but if the two sine waves are
not in the right phase relationship, there might not be any power consumed.
Strange as that seems, 1000 amps at 1000 volts with an angle of 90 degrees
between current and voltage doesnt consume any 'direct' power, but the
current will cause a 'real' voltage drop in the wiring, which will still
use some power. Of course this means that the actual phase angle wouldnt
be exactly 90 degrees then, but maybe close to it, like 89 degrees.


Good luck with your 'ac' circuits

Al