Re: power factor correction

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Jim Lux" <jimlux@xxxxxxxxxxxxx>


----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Saturday, February 12, 2005 10:16 PM
Subject: Re: power factor correction


> Original poster: Dan Kunkel <dankunkel@xxxxxxxxx>
>
> Gerry,
>
> allow me to play devil's advocate here...
> if you are reducing line current and not tripping breakers because of
> your PFC's, then it seems to follow that you are in fact using less
> power. maybe an explaination of true power is needed.
>
> Dan
>
>
Sure enough, Gerry is right.

Power is the average of the instantaneous current times instantaneous
voltage.  If you hook a resistive (non-reactive) load across the line, you
could measure the voltage and current with (rms) meters, multiply them and
get the power. P(t) = E(t)*I(t).. However, that's a special (but very
common!) case, because the current and voltage are exactly in phase.

 If, for example, you hook a perfect (lossless) capacitor across the AC
line, current will flow, but there will be no power.  The current will be 90
degrees out of phase with the voltage, so when you average out
(sin(2*pi*60*t)*cos(2*pi*60*t)) you get zero.

If you hooked up an ammeter and voltmeter, they'd both read something, but
multiplying them isn't going to give you the power (like it would with a
non-reactive load or DC). It would give you VoltAmps.

The current's still important, because in any of the resistive parts of the
circuit, it would still dissipate heat.  (why transformers, for instance,
are rated in VoltAmps, or more commonly kiloVoltAmps (kVA))

The electric meter on the side of your house is a very clever analog
computer that essentially does the multiplying and integrating on the fly
(it has two coils, one for current, one for voltage, and the two coils form
a 2 phase induction motor with the aluminum disk, where the disk spins at a
speed proportional to the product of the two magnetic fields)

http://home.earthlink.net/~jimlux/hv/pfc.htm has some more info

> On Sat, 12 Feb 2005 22:46:52 -0700, Tesla list <tesla@xxxxxxxxxx> wrote:
>  > Original poster: "Gerald  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
>  >
>  > Hi Dan,
>  >
>  > I dont think the PFC will effect the electric bill since the KWH meter
>  > measures true power.  It will reduce the line current and help keep
>  > breakers from tripping for coils that draw high enough power.
>  >
>  > Gerry R.
>  >
>  > .
>  > >Original poster: Dan Kunkel <dankunkel@xxxxxxxxx>
>  > >
>  > >PFC's are NOT required, they just help to bring the voltage and
>  > >current in phase with each other, there by reducing total current draw
>  > >from you electric company and lowering you electric bill. this is
>  > >important for home air conditioners, fridgerators etc...
>  > >
>  > >hope this helps,
>  > >Dan
>  >
>  >
>
>