# Re: Pig Ballast Question (fwd)

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---------- Forwarded message ----------
Date: Sun, 12 Jul 1998 04:05:53 -0600
From: David Dean <deano-at-corridor-dot-net>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Pig Ballast Question (fwd)

-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com <tesla-at-pupman-dot-com>
Date: Sunday, July 12, 1998 1:56 AM
Subject: Pig Ballast Question (fwd)

>
>
>---------- Forwarded message ----------
>Date: Fri, 10 Jul 1998 22:17:00 +-100
>From: "Gregory R. Hunter" <ghunter-at-enterprise-dot-net>
>To: 'Tesla List' <tesla-at-pupman-dot-com>
>Subject: Pig Ballast Question
>
>Dear List,
>
>A fellow coiler told me that if I use a purely resistive ballast, I'll
only get about half the rated power out of the pole pig.  In other
words, if I use a 5KW space heater as a ballast, the pig will only yield
about 2.5KVA output.  Any truth to this?  Sounds screwy to me.
>
>Greg
>
Yes, this is true.

Assume a 5KVA transformer with a 5KW heat strip in series with the
primary. (240 volt side).
(The 5KW heater is a "pure" resistance so phase shift is "0", so KW =
KVA.)
When the transformer is unloaded the "pri." will pull about 2 amps or
so, so most of the voltage will
be dropped across the transformer pri. and very little across the
resistor, so output voltage will be almost
rated voltage. When the transformer is fully loaded the current draw
goes up, the drop across the resistor
goes up till it equals almost the input voltage, and all the power is
dissipated in the resistor as heat.
Maximum power transfer will be when the loading on the transformer is
such that its impedance as seen by the power supply is exactly equal to
the resistance of the heater, that is the input voltage is divided
evenly between them. One half of the power is dissipated in the heater
as heat, one half of the power is delivered to the transformer to do
work. A nice way to make a pole pig coexist with a 30 amp circuit
breaker.

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