Capacitor wiring? PF - amperage reduction

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Craig_Karnitz

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Gentlemen Hi,
I have a problem I hope one or two of you can explain and help me
understand. A few weeks ago a local businessman here on our island
approached me to set up a demo- display for a new product he was
going to sell. (our utility company is charging us .34 cents per kwh now
and July 1st it will increase to .42 cents ?ouch? People here are looking
for miracles) He showed me a internet video of the display, and I said,
sure I could set one up. I wired it up as per supplied instructions, code,
and common practice. I supplied the capacitor unit from a dedicated 2
pole 20 amp breaker. I installed 2 ? Killawatt meters to view the reported
reduction in amperage and the improvement in power factor.
After powering the unit up I as unable to duplicate the savings seen in
the video. After trying to contact their technical help person for a week
and a half, I finally found another of their distributors that sent me their
?magic wiring diagram?
Basically the only difference from the circuit I drew up and theirs, is that
they wanted the display circuit to be fed from a single pole (120v) breaker.
L-1 on their unit to the breaker and L-2 to the neutral.
WELLA!
I can now see the reduced amperage 4.3 amps down to 1.5, and the
Improvement in power factor .23 prior to turning the motor on ? to .83
with it running????
It has been a long time since I had to work with this in class!
I?m just an ?older? electrician, but this is looking like smoke and mirrors to me?
Can one of you with an engineering background explain how the wiring change
of how the capacitor discharges into the circuit 240 to 120 is affecting this.
Is it legitimate use to change the circuit wiring to see these magical savings?
Thanks in advance for any help!

Craig
 
080624-0822 EST

Craig:

If your meter is measuring integrated power (KWH) and nothing else, then for the most part you do not care what the power factor is.

All resistive loads have a power factor of 1 anyway. Incandescent lamps, electric heaters, flat irons, electric dryers other than the motor, waffle irons, ovens, stoves, hair dryers, and so on are all basically resistive loads. A mechanically unloaded single phase motors may have a somewhat low power factor, but a mechanically loaded capacitor run motor has a pretty good power factor. How many motors do you have that run with virtually no load and how often do these run?

None of this matters much if your meter only reads power and power factor is not part of your billing. The integration (summation) of power over time is the energy you use and for which most residential customers pay.

Save money by measuring your base load that is on 100% of the time and eliminating unnecessary stuff. Like always on TVs and accessories, computers, and anything else you can find. My base load is in the 100 W range. Some of this is dimmed lights I never turn off. At 8760 hours per year that is 876 KWH. Last year my rate was $0.1178 per KWH and my base load cost about $103. For your new rate and a 100 W base load your cost would be 0.42*876= $368. Thus, reducing your base load to an absolute minimum may be of great benefit. Furthermore maybe consider motion detector switches for lights, and possibly with additional logic.

.
 
gar said:
Some of this is dimmed lights I never turn off.
I once made a bright/dim (but not off) for the hall-bath light by placing a diode across the switch. It was a 60w bulb, and I used a 1a 400v diode (1N4004, if memory serves.)
 
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