Re: Looking for run caps, urgent

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

But folks, you are calculating based on a purely inductive load (i.e. the
entire VA is VAR).  In a real system, you're going to have some resistive
component, which greatly helps the power factor computation.  The power
factor is the ratio between the active power and the apparent power (or,
alternately, the cosine of the phase angle...).  A power factor of 0.85,
for instance, means that for every 1000 VA you measure as volts times amps,
you're actually using 850 Watts.  In this particular example, there would
be about 527 VAR of reactive power.

Here are two equations:
Apparent power^2 = activepower^2 + reactivepower^2
power factor = activepower/apparentpower

Apparent power = what you get when you measure with an ammeter and voltmeter
Active power = what you get when you measure with a real watt meter

With a purely inductive load, your PF is going to be 0, regardless of how
much capacitance you hang on there, because there isn't any active power
for the numerator of the fraction (active power/apparent power)

As a practical matter, consider why are you doing PFC in the first place.
I can think of three reasons:

1) The power company wants your load to be reasonably high power factor. I
doubt this is the case for any tesla coiler, except perhaps in a large
permanent installation.
2) You want to reduce the line current being drawn for a given amount of
real power being consumed, so you don't trip the breakers, overheat the
variac, etc. - This is the real motivation, because it means that you
aren't getting limited by anything other than your transformer size.
3) You think it is technically elegant to have good PF - go for it, but
recognize that it's in the same category as using finely polished burlwood
on the control panel. Very nice, shows attention to detail, but doesn't
have a heck of a lot of effect on performance.

For #2, you just need to figure out how much inefficiency (headroom), etc.
you're willing to tolerate.  And remember that this is only at full load.
At low loads, who cares if it is inefficient. 85% is a nice working number.  




Tesla list wrote:
> 
> Original poster: "Michael Rhodes by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <rhodes-at-fnrf.science.cmu.ac.th>
> 
> I ran the calcs through my program and come up with the
> following:
> 
> 15kV-at-60mA 220VAC/50Hz = 59.19uF
> 15kV-at-60mA 110VAC/60Hz = 197.3uF
> 
> 15kV-at-120mA 220VAC/50Hz = 118.38uF
> 15kV-at-120mA 110VAC/60Hz = 394.6uF
> 
> --Michael
> ----- Original Message -----
> From: "Tesla list" <tesla-at-pupman-dot-com>
> To: <tesla-at-pupman-dot-com>
> Sent: Tuesday, October 23, 2001 1:26 AM
> Subject: RE: Looking for run caps, urgent
> 
> > Original poster: "Ted Rosenberg by way of Terry Fritz
> <twftesla-at-qwest-dot-net>" <Ted.Rosenberg-at-radioshack-dot-com>
> >
> > Jonathan:
> > There seems to be some weirdness here.
> >
> > Terry...you might want to jump in.
> > I ran the formula using my 15/60 as a model. The answer was 165µF.
> > But in an earlier e-note you suggested that only 100µF would be enough.
> >
> > Now here's Jonathan with a figure that meshes with my 165, ie 165 * 2
> > <because he's using a 120mA figure>= 330.
> > So, what's the REAL story?????
> > Inquiring minds want to know  :))
> >