Date: Tue, 12 Mar 2013 11:34:03 +0000
From: David Kirkby <david.kirkby@onetel.net>
To: jeff millar <wa1hco@wa1hco.net>
Cc: amps@contesting.com
Subject: Re: [Amps] Freescale LDMOS devices
On 9 March 2013 04:34, jeff millar <wa1hco@wa1hco.net> wrote:
> Using a power factor correction cap works when the load is a bunch of
> inductive motors. But it won't really work for the pulse type load from a
> capacitor input filter.
>
> Power meters measure RMS current and RMS voltage. But Watts RMS only equal V
> rms x I rms when the power factor is 1. For lower power factors you pay for
> Vrms x I rms which is greater than Wrms.
I did not finish composing this before Google decided to send it. So I
will start again.
RMS power is not a useful quantity. Whilst it is possible to compute
the RMS value of power if you want to do the maths, it is of no
practical significance in any application whatsoever. Whilst RMS
voltage, RMS current and mean power all have their uses, RMS power
does not. This is irrespective of what the application is - i.e. it is
not specific to power factor. As such, as someone else noted, he knows
of nothing which measures RMS power. That said, if would not surprise
me if you found some cheap meter from a mickey-mouse manufacturer what
claims RMS power, or you find power amplifiers rated in terms or RMS
power, but it is still a useless quantify.
For pure sine waves of Vrms and Irms, the mean power is the same as
the DC heating effect if I and V were contant (i.e. DC). The RMS power
is not.
Whilst I don't know how your particular electric company bills you, I
very much doubt it is purely by Vrms * Irms. Most will bill by the
mean power, which is
V * I * cos(theta)
where theta is the angle between the voltage and current.
There are two issues with having a low power factor.
1) The wiring in your house will be carrying more current than needed
to supply the same mean power to you load. So the resistive losses in
your own wiring will mean that end up paying more than you want to, as
more power will be wasted in the resistive losses in your cables.
2) If you are a large user of electricity, the power company will not
permit you to run at a low power factor. One obvious impact on them is
that the resistive losses in the cables reaching your property will be
higher for the same amount of power they charge you for, so they are
not keen on it. I suspect their generators would have issues supplying
very reactive loads too. I don't know much about that, so I'll comment
no further.
For a small user of electricity, I suspect the cost involved in trying
to correct the power factor of reactive loads will outweigh the
benefit you will gain from slighly lower resistive losses in the
cables. I guess for a QRO amp, you want as much power as possible, so
correcting the power factor will reduce the resisitive losses of your
wiring and give you a slightly higher power output. But it is not
going to make a huge difference.
Dave
## and how is the typ power meter for a residential home going to be able to
know what the PF is ?? I’m sure for residential metering it’s just Vrms x
rms amps
= VA power. So if you have lousy PF, you end up sucking more VA power,
and you pay more.
## If you are indeed correct, and the power co meter reads rms power, and not
VA power,
then the power company is being screwed if joe home owner happens to have a
lousy PF
cuz of some huge inductive load, like ac motors..... or joe ham with a high C
HV supply, also
with lousy PF. On ssb/cw there is no way to correct for lousy PF, when high C
filter caps are used.
## use bigger ga wire back to main panel if using a high C HV supply. Then
you get less V drop,
so now you can suck even more power. You are not doing yourself any favours
using a long run of 12 ga
wire from shack to main panel. To reduce clutter, I ran 3 ga from main 200A
panel, to a 100A sub panel,
on other side of shack wall. Then all the runs from sub panel to shack are
extremely short. Even then I still
used 10 ga wire from sub panel to each table top amp.
Jim VE7RF
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