response 26 of 83:

Mar 28 12:41 UTC 1999

Heck, I don't really understand it myself.  Anyway, it is not an exact science
when applied to commercially available UPSs.

response 28 of 83:

Mar 28 23:35 UTC 1999

Ok, Rane, since you volunteered:  What's a capacitor?  What do you mean when 
you say "VA is the average product of the voltage and current if they are *in
phase*" - in phase how?  I understand that you're saying that current is
proportional to the derivative of voltage, so that if
        voltage = a*sin(wt)
for some constants a and w, then
        current = b*cos(wt)
for some constant b.  Then
        power = voltage * current = a*b*sin(wt)*cos(wt) = (ab/2)sin(2wt)
which indeed averages 0 over a long stretch of time.  But why is "VA"
different from power?

response 30 of 83:

Mar 29 06:43 UTC 1999

A capacitor consists of two conducting plates (or films) in close
proximity but not touching. When a voltage is applied an static electric
field is set up in the space between in which energy is stored. If you
remove the voltage source, the static field remains as does also the
voltage difference between the two plates. If a conductor is then put
between the two plates, the capacitor discharges as the field is drained.
This all occurs even if there is a vacuum between the plates. If some
material is put there, one not only gets better insulation between the
plates, but the atoms of the material participate in the field and
displacement of electrons and nuclei occur, which considerably increases
the capacitance.

If both the current and voltage are in phase, we can write v = Vsin(wt) 
and i = Asin(wt), where V and A are the maximum voltage and current
respectively. Power (work per unit time) is the product of voltage and
current at each instant. If i and v are in phase, this product is always
positive, and work is being done (whether in a motor, being converted to
mechanical work, or in a resistor, being converted to heat). If i and v
are out of phase less work is done, since i*v is alternatively positive
and negative, where if it is negative it means that the system (capacitor) 
is doing work hurrying those little electrons along. If we call the
leading phase shift between i and v p, v = Vsin(wt) but i = Isin(wt + p). 
So when p is not zero, the circuit is alternately absorbing energy and
releasing some of it back into the circuit, until when p = pi/2 (radians =
90 degrees), the absorbed and returned energy are identical and no work is
done. (Look upon the condition of the current flowing in the opposite
direction with voltage in the positive direction as the system being a
generator.) 

response 32 of 83:

Mar 29 06:57 UTC 1999

Power = push x speed.  In electrical terms, push = voltage, & speed =
current.  For DC work, it suffices to measure v & i (current), & compute
the power directly.  With AC, however, the voltage is in fact constantly
varying; so what you really need to compute is the integral of the
product, v*i, computed across one complete cycle.  If the load is purely
resistive, i is directly proportional to the instantaneous v, & this is
easy.  A capacitive or inductive load moves the "i" curve so that it is
out of phase with the "v" curve.

A pure capacitor takes no current at constant voltage, and only eats
current if the voltage changes.  Basically, a capacitor "stores" a
charge.  An inductor "stores" current - it will "push" back (and
generate a back voltage), if something tries to push more current
through, or resist sending as much current through.  If you were to
compare electricity to water; voltage = water pressure, current = water
flow, a capacitor is a big rubber diaphram that can "store" water, and
an inductor is a big length of pipe with really smooth sides.  With
water, if you shut off a valve very quickly, the water that's flowing
through the pipe to the valve has momemtum in it, and will try to "kick"
at the valve (and generate a pressure spike) as it tries to keep flowing
- this is the water hammer effect.  This works with electricity as well
- the ignition coil in an automobile is basically a really long slick
pipe to the electricity, and when the contacts in the distributor open
(assuming your car is that old), the electricity that had been flowing
through the ignition coil generates a voltage spike, and jumps the gap
in the spark plug instead.

So, getting back to AC.  Suppose V is a sine wave.  At "0", the voltage
is rapidly increasing, so lots of current is flowing into the capacitor
to charge it.  But since V is near 0, there is almost now power.  At
pi/2, the voltage is at a max, but not changing much at all.  The
capacitor is fully charged with energy, but again, little power is
flowing.  At pi, the voltage is dropping rapidly, and lots of current is
flowing, but again, voltage is near 0, so there is little power involved
here.  At 3*pi/2, v is near -1, & i is near 0, and so the cycle goes.
Inbetween the quadrants, v & i are both non-zero, so power is in fact
happening, but half the time, this power is positive, & half negative.
Basically, the current is in advance of the voltage curve by 90 degrees,
and the power that is generated in 1/2 of the cycle is consumed in the
other half, so no real work is generated.  With an inductive load,
something similar happens except that the current lags behind the
voltage by 90 degrees.

If you combine an inductor and a capacitor, you can cancel out the two
effects, and end up with a purely resistive load instead (if the two are
"=", which in turn depends on the frequency and the relative values of
each).  In the water model above, that would be equivalent to putting
some sort of air filled chamber near the valve that's being shut off.
When the water gets turned of quickly, the water that's flowing will
compress the air in the chamber instead of trying to break the pipe.
This is sometimes done with washing machines, for instance.

response 34 of 83:

Apr 1 00:59 UTC 1999

I just read the preceding 10 responses or so to Jim, pant, pant.
Who says "I am really concerned what my electrical meter thinks the difference
is between VA and W.    I don't think my electrical meter cares. If I ran
half-wave load off my electric meter I think it's still going to give me a
full reading of usage.  If they say there's a difference, what effect does
the difference have on the average Joe or Jane in the street or in the
kitchen?  For DC VA=W, for AC it averages out that way over the long run?
If it's not, I want to know what kind of load to put on my electrical meter
so that I can receive the most for my money.  Which appliances run more
efficiently - resistive, inductive, capacitive."  (What's a capacitive
appliance, Sindi asks.  Jim says 'a radio'.  Inductive appliance - motor. 
Resistive - stove.).

response 36 of 83:

Apr 1 18:51 UTC 1999

        P= IE
        E= I^2 R

        Nuff' said.

response 38 of 83:

Apr 2 13:27 UTC 1999

        <shrug> Not according to the texts and labs I had at WCC, but..
        "oh, well"..

response 40 of 83:

Apr 13 06:05 UTC 1999

Your electric meter measures va*h, not w*h (despite its name of
"wattmeter").  If your power factor is way off, then essentially you get
less energy, or the energy you do get is way more expensive.  This is
the power company's way of encouraging you to have a more sane power
factor.

For grex, the difference is probably not that important.  When sizing a
UPS, you don't get a unit that is "exactly" big enough, you get a unit
that is significantly over-sized.  That way, adding an extra board, or
having components that age and change ratings, won't cause capacity
problems.

response 42 of 83:

May 16 03:44 UTC 1999

Mebbe this should be posted in the 'random grexer sightings' item...

This afternoon, I observed a Grex board member finalizing a deal for a
1.5KVA Liebert UPS.  I'll let him post the details, But I'll say it was a
fantastic deal!

response 44 of 83:

May 18 10:19 UTC 1999

Will this unit run in the pumpkin without changes to our
electric service?

response 46 of 83:

May 18 20:34 UTC 1999

(You opened resp:45 with "No", but from the rest of your response it
sounds like you meant "Yes".)

response 48 of 83:

May 18 23:04 UTC 1999

Mary's question may have been related to rewiring, adding a 220 V outlet, etc.