Home PF Correction?

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wptski said:
They were TCPI. In fact, I had to return one of a two pack that didn't work! The girls asked where the other one was and I said in a socket, working. She said that I had to return both of them!:rolleyes: Ended up, splitting up a two pack.

Here's where, I think that 33W for 18W CFL is mentioned.
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Let's ignore Jraef's not about displacement pf for a moment and have a sanity check:

Let's take a 2 mile, 3-phase, #2 ACSR, 7.2 kV feeder and peak load it to 1701 kW at a 60% pf and 100% load factor. We have annual kWh of 14,900,760 and annual losses of 1,276186 kWh (816,759 kWh due to var). This is a 8.565% loss. Let's give another 4% for generator to the feeder source and another 5% for the transformer and service drop and 3% for the home.

So even accounting for all of these losses, we only get to 22.7W and that is not all because of var.

I just don't believe the 33W number.
 
It isn't 33W, it is 29VA, and therein lies the difference.

mivey's referenced article said:
...so a 15W CFL will actually draw just under 29VA.
Click to expand...

But let's look at this a little differently. Assuming that if we are looking at a 15W CFL and that figure is correct (even the writer doesn't say that for sure), then to get the equivalent light output of a 75W incandescent bulb, we are going to demand 29VA of power from the transformer. IT IS STILL HALF OF WHAT IT WAS! When it was a 75W incandescent bulb, it was demanding 75VA from the transformer! How then is this an additional "burden" on the transformer (or even the generating facility) that needs to be worried about?

Bottom line; sure, power factor improvement helps the utility, but they are NOT charging us for it in residences, so why bother correcting it? Unless we all go crazy and say that we want our houses to be twice as bright on the inside as they used to be and put in twice as many CLFs as we had incandescents, this is not a problem for the utility at all, it is still a good thing (from an energy standpoint anyway).
 
080701-1902 EST

I took a look at the following article from an above reference.
I think it was wpstki's #16 and the link TCPI but I can not find the path to the following web sites from TCPI site.

http://sound.westhost.com/articles/incandescent.htm
power factor tab

http://sound.westhost.com/articles/incandescent.htm#pf
Figure 11 waveform and fig 6 has a circuit.

I just ran an experiment with a Commercial Electric SKU 722-429 Model EDXO-19LPF, 19 W 0.208 A 120 V 60 Hz. I get a waveform very much like figure 11. My peak current was 1 A. The start of the sharp rise is at about 2.5 MS and the trailing end about 5.0 MS. 2.5 MS is about 53 deg and 5 about 106 deg. Thus, this has a leading power factor. A capacitor does not compensate this. There is a lot of high frequency content in this waveform compared to 60 Hz.

By contrast my Tektronix 7603 has about a 3 A peak is relatively centered at about 90 deg the peak of the voltage waveform. It has a broader base, is smother, and more like a narrowed half sine wave.

A mid 90 IBM Desktop computer has about 2 A peak and is similar to the Tek scope except somewhat narrower.

If the peaking of the CFL is as bad as indicated then there can be real problems with a large number of these on a system. To the extent that the distribution system near the point of use can be made into a low pass filter the problem back toward the alternators won't be so bad.

I use some compact fluorescents and they are very useful for an always on light. But mostly I use 8' Slimlines. I need to look at my current waveforms and see what typical usage is like.

But relative to post #1 the device is still a fraud and would not do much to correct CFL power factor other than as part of a low pass filter and it might create greater power factor problems if the capacitance was large enough to do much filtering. I do not know just guesses here.

In a rectifier circuit if you make the series inductance sufficiently large relative to the load current, then you get current conduction thru the full cycle. This is known as a choke input filter as distinguished from a capacitor input filter. Heavier and more expensive than a capacitor input filter. Used in the early days of radio to serve as the electromagnet for the speaker. This was before the electrolytic capacitor made large capacitors inexpensive. Choke input filters require much smaller filter capacitors for the same ripple.

There are people suggesting a dual power supply. 120 V AC, and DC supply of comparable or higher voltage. If much of the load was on a DC supply, then the interface to the AC line could probably provide near unity power factor. I did not like the idea of sharing the receptacle. Their claim is that fluorescent lighting will be more efficient and reduce the power dissipation in the fixture using DC. The overall efficiency may not be better, but the fixture might be, and a lower temperature would be advantageous.

.
 
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080702-0658 EST

An interest point of one of the links I listed above was this analysis:

If you are at a 50% power factor this means the input RMS current is double the current of the work being done. Since the power dissipated in the distribution system is I^2*R there is a 4 times greater distribution system power loss at the 50% power factor. This extends into the alternator.

System design has to be based on maximum temperature rise and approximately a 4 times greater temperature rise will occur. This means all the distribution system including the source has to be about 4 times larger in power capability than for unity power factor.

This does not mean the energy input to the system is 4 times greater. It does mean the actual input is the load energy + 4 times the system losses. If the system losses were 5% at unity power factor (and I have no idea what they are), then the 50% power factor load system losses would be 20%.

An important question is to what extent, and how quickly in terms of distance, is the messy CFL waveform modified as it moves back thru the distribution system?

.
 
gar said:
080701-1902 EST

I took a look at the following article from an above reference.
I think it was wpstki's #16 and the link TCPI but I can not find the path to the following web sites from TCPI site.

http://sound.westhost.com/articles/incandescent.htm
power factor tab

http://sound.westhost.com/articles/incandescent.htm#pf
Figure 11 waveform and fig 6 has a circuit.


.
Click to expand...

Thanks for the reference, that was interesting to read through. I always enjoy seeing waveforms.

You do get into it don't you ;)
 
gar said:
080701-1902 EST

I took a look at the following article from an above reference.
I think it was wpstki's #16 and the link TCPI but I can not find the path to the following web sites from TCPI site.

http://sound.westhost.com/articles/incandescent.htm
power factor tab

http://sound.westhost.com/articles/incandescent.htm#pf
Figure 11 waveform and fig 6 has a circuit.
Click to expand...
Nope, there's no link from TCPI's site to those two. I provided two seperate links.

Try this one, scroll down to the last two entries for some Pro's/Con's.
 
080702-1439 EST

The CFL that I tested in post #23 has very poor dimming characteristics using a Powerstat (Superior Electric's name for a variable autotransformer).

Between 90 and 120 V intensity is fairly constant. Intensity changes somewhat going down to 65 V. At about 65 V the heaters (filaments) in the fluorescent get too cool and proper operation ceases.

.
 
I purchased one of these dimmable CFLs :

http://www.smarthome.com/903233.html

Spendy huh?

Because they claimed to be very dimmable. So far I have been happy with it. I have a lot of home automation dimmers and was curious. I probably will not buy anymore at that price.
 
Their specs didn't show it. I bet Gar could test it though if someone were to willing to donate a sample :smile:
 
I don?t understand this power factor stuff very well at all.
So bear with me for a stupid question.
I get free lunches all the time so I can?t dismiss it on the ?no free lunch? theory.
I do know that on the residential side our power company does not adjust the bill for power factor on an individual basis.
So correcting the power factor closer to ?1? is not going to reduce the bill.
As I understand it the meter only measures current and voltage that are in phase with each other to calculate the watt-hours.
So my question is: if your home were made to have an artificially poor power factor, would any of the resistive loads in the home dissipate more power that the watt-hour meter does not ?see??
 
johnny watt said:
I don’t understand this power factor stuff very well at all.
So bear with me for a stupid question.
I get free lunches all the time so I can’t dismiss it on the “no free lunch” theory.
I do know that on the residential side our power company does not adjust the bill for power factor on an individual basis.
So correcting the power factor closer to “1” is not going to reduce the bill.
As I understand it the meter only measures current and voltage that are in phase with each other to calculate the watt-hours.
So my question is: if your home were made to have an artificially poor power factor, would any of the resistive loads in the home dissipate more power that the watt-hour meter does not “see”?
Click to expand...
If you dissipate power in your home, the meter will see it.

Let's consider that the losses in your home don't matter for a moment. Let's give you the worst power factor in the world, which be a totally reactive load. Your meter will show no usage.

Reactive load in the home will cause small heat loss and small voltage/current changes. Within the voltage levels, loss levels, and percent reactive load you see in a home, you will never save enough to pay for this device, or even an appreciable part of it, even if this were put on a subpanel.
 
johnny watt said:
As I understand it the meter only measures current and voltage that are in phase with each other to calculate the watt-hours.
Click to expand...
You do get it, mostly. The meter will see 100% of the resistive load. and a percentage of the reactive loads (100% of the resistive portions).

The meter is a motor whose speed depends on the voltage and the current (watts). When the voltage and current peaks are offset in time, the motor runs a tad slower, yet the full voltage and full current still have to be carried and delivered by the system.
So my question is: if your home were made to have an artificially poor power factor, would any of the resistive loads in the home dissipate more power that the watt-hour meter does not “see”?
Click to expand...
No, the resistive elements will not see, nor produce, any changes with additional low-PF loads. Resistive and reactive components add in a system. Both types add to produce the total. Adding low-PF loads will not 'disguise' the unity-PF currents.
 
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Maybe the poco has made measurements of PF on different distribution circuits and has a typical value that they take into account when setting the rates. Residential power bills confuse enough people right now because of transmission, distribution, generation, and cost to compare charges. How would the average consumer respond when half the electricians and EEs can't accurately describe W vs. VA vs. VAR and what it really means? Just because it isn't itemized or measured on a small scale doesn't mean you aren't paying for it and that it isn't measured in aggregate for larger areas.
 
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