Re: Power factor (was genset trailer)

[Lee Hart <leeahart@earthlink.net>]

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