Re: pole pig chart (fwd)

---------- Forwarded message ----------
Date: Fri, 24 Apr 1998 14:11:32 -0700
From: Jim Lux <jimlux-at-earthlink-dot-net>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: pole pig chart (fwd)

> ---------- Forwarded message ----------
> Date: Thu, 23 Apr 1998 23:39:57 -0500
> From: Jeff Corr <corr-at-enid-dot-com>
> To: tesla list <tesla-at-pupman-dot-com>
> Subject: pole pig chart
> I am just curious...
> does anyone have a chart showing the relationships
> between kVA of the pig, with the current it pulls on
> the 220 side if it has an output of 13.8kv or 14.4kv for
> Tesla Coil use, ex 

No chart needed:
Use Ohm's law:

Amps = 1000 * kVA/ Voltage

example: 10 kVA

1000 * 10 / 240 = 41.7 Amps

1000 * 10 / 13800 = 725 mA


Amps = kVA / kV

VA is just VoltAmps, which is the volts multiplied by the amps.

Transformers are not rated in Watts because the current may be out of phase
with the voltage, reducing the actual energy flow (i.e. the Watts), but
still having the same Volt Amps (aka Apparent power).  The losses in the
transformer are generally due to the Amps, not the Watts, so the losses are
a function of the load's VoltAmps, not the load's actual energy

For example imagine a big 1000 microfarad (lossless) capacitor across the
240 Volt 60 Hz power line. If you hooked an ammeter up, you'd see a current
of 90 Amps flowing. (The reactance is about 2.66 ohms) However, no work is
being done, and the capacitor isn't getting hotter (it is lossless,
remember). This circuit has an apparent power of 21,600 VoltAmps (or 22
kVA), but an actual power of ZERO Watts. Your electric meter wouldn't even
move if the wiring were also lossless.

However, all those amps do make the wires warmer, due to the I^2*R losses.
The 90 amps will also trip the breaker, since the breaker looks at the
current, not the power.

And, if you were an industrial consumer, and your distribution transformer
overheated, even though your electric bills were low, the utility would
come out and take a look, figuring that either you had bypassed the meter,
or that you had a "BAD POWER FACTOR", and would start charging you for the
kVA in addition to the kWh.

In reality, most industrial consumers, with a lot of lightly loaded
induction motors, have a net inductive reactance, not a capacitive
reaactance, like in my example. The utility (or the consumer) compensates
for this inductive reactance by adding "Power factor correction"

more at http://home.earthlink-dot-net/~jimlux/hv/pfc.htm