Re: Power factor (was genset trailer)

["1sclunn" <1sclunn@msn.com>]

Steve Clunn wrote
thank you Lee for explaining what I was seeing here.  I stopped
experimenting with this when I though it was wasting 1/2 the power.  what
about combining both non-electrolytic "run" capacitors charger with a
isolation step down  transformer type charger with both in parallel on the
240 ac to give a 120v charger and  use   snip; a relay coil with zeners or
neon lights in series to sense battery voltage. When it gets high enough,
the relays pulls in, and its contacts
disconnect some of the capacitors.If both were putting out the same amps
would the lead and lag = out ?  until I can afford a pfc in need to make the
best substitute

----- Original Message -----
From: "Lee Hart" <leeahart@earthlink.net>
To: <ev@listproc.sjsu.edu>
Sent: Wednesday, July 31, 2002 2:44 PM
Subject: Re: Power factor (was genset trailer)


> 1sclunn wrote:
> > I hooked up some capacitors that way with a bridge rectifier in place
> > of the light bulb and some batteries on the dc side of the bridge.
> > I was surprised to find that when the input voltage went up the
> > current went up but not like it would when you raise the voltage on a
> > battery being charged.
>
> That's right. The capacitor (or inductor) acts like a series resistor to
> limit the current, so raising the voltage does not raise the current as
> much. However, a resistor converts the excess energy into heat. A
> capacitor or inductor do not get hot; they shift the phase (lower the
> power factor) which does not produce heat.
>
> > I was charging my 120v car on 240v(ac) and was controlling the amps by
> > adding caps (air-conditioned motor caps).
>
> Hopefully, you used non-electrolytic "run" capacitors, not the
> electrolytic "starting" capacitors. Electrolytics have a lot of
> resistance, and will get hot and fail (dramatically and messily)!
>
> > 1. The amps flowing into the car (on E-meter) were the same as at the
> >    pole... were did the other half of my 120 go?
>
> Yes; it was very observant of you to notice! The current in (from the AC
> line) has to be the same as the current out (DC into batteries), since
> there is no transformer or other converter.
>
> As for where the voltage went; half of your 240vac was across the
> battery pack being charged. The other half was across the capacitor.
> This capacitor was carrying 120vac and the full charging current. But it
> doesn't get hot because the voltage and current are 90 degrees out of
> phase.
>
> > 2. The thing went right by 150 volts, this is where it usually would
> >    taper off. When I checked on it the voltage was 160 and 8 amps
> >    about.
>
> That's because, in effect, you were charging the pack from 240 volts
> thru a "reactance" of (240v-160v)/8amps = 10 ohms. Changing the battery
> voltage has little effect on charging current because it is small
> compared to the AC line voltage.
>
> If you had been charging on 120vac, then battery voltage is large
> compared to AC line voltage, so small changes in battery voltage have a
> big effect on charging current.
>
> > 3. the lights in the shop seemed brighter when I plugged it in
>
> That's possible, if the rest of the AC loads in your shop were inductive
> (motors, transformers, old-style fluorescent lights). The capacitor was
> drawing a "leading" current, while everything else was drawing "lagging"
> currents. The leading current cancelled all or part of your lagging
> currents, improving your shop's overall power factor. This has the
> effect of reducing your total AC line current, and thus reducing voltage
> drop, and raising your line voltage. But the effect is quite small.
>
> > I don't under stand exactly what was happening. Was I feeding back to
> > the grid the same amps as the car was seeing just out of phase?
>
> Sort of. Power factor is hard to explain because there are no good
> analogies. But you are seeing its effects quite clearly!
>
> Try this: Suppose you are trying to push a heavy box across the floor.
> You can't do it by yourself, so you call your brother-in-law Homer to
> help. You yell "push" and shove for all you're worth. Homer says,
> "When?" and just stands there. The box doesn't move. You stop pushing
> and say, "Now, stupid!" So he pushes, and you stand there. The box still
> doesn't move. You are both pushing as hard as you can, but no work is
> being done. You two are working "out of phase".
>
> So you say, "Push when I say 'now'". You push and say "Now". He pushes
> at the same time, and the box moves. Now you are working "in phase".
>
> Power is Volts times Amps. But in AC circuits, the volts and amps do not
> flow continuously; they pulse on and off, many times a second. If the
> voltage happens to be zero when the current is present, 0 x Amps = 0 and
> there is no power. If the current happens to be zero when the voltage is
> present, Volts x 0 = 0 and there is still no power.
>
> This is what goes on with a capacitor or inductor. The volts and amps
> are still there, but not at the same time; the power is zero.
>
> Your capacitive charger idea will work fine, but you need some way to
> sense battery voltage and turn it off when fully charged. For example,
> use a relay coil with zeners or neon lights in series to sense battery
> voltage. When it gets high enough, the relays pulls in, and its contacts
> turn off the charger, or disconnect some of the capacitors.
> --
> Lee A. Hart                Ring the bells that still can ring
> 814 8th Ave. N.            Forget your perfect offering
> Sartell, MN 56377 USA      There is a crack in everything
> leeahart_at_earthlink.net  That's how the light gets in - Leonard Cohen
>
>