On Fri, 29 Mar 2013 12:59:34 -0700, Gary Walters
wrote:

In USA.

Source: 208v, 60 hz, 2-wire (2 phases from 3 phase "Y" supply). Load: 240v,
20A.

I presumed that sizing a buck-boost transformer is simple KVA math (source
volts * load amps). But...

This PDF document:

http://www.acmepowerdist.com/pdf/Page_104-109.pdf

on the last page says:
- - -
"An example of an everyday application is always a good way to explain the
intent of the ³Code.² Example: A 1 kVA transformer Catalog No. T111683 has a
primary of 120 x 240V and a secondary of 12 x 24V. It is to be connected as
an autotransformer at the time of installation to raise 208V to 230V single
phase.

When this 1 kVA unit is connected as an autotransformer for this voltage
combination, its kVA rating is increased to 9.58 kVA (may also be expressed
as 9,580 VA). This is the rating to be used for determining the full load
input amps and the sizing of the overcurrent protect device (fuse or breaker)
on the input.

Full Load Input Amps =
9,580 Volt Amps / 208 Volts = 46 Amps"
- - -
I'm puzzled by the 10x increase of KVA rating. When and how is this true?

What size B-B transformer do I need?

Thanks.

The secondary of an autotransformer carries the load current but only
supplies the voltage *difference*. 240-208 is only 32 volts, so at 20
amps the boost secondary is delivering 640 VA.

True *if* the boost transformer is being used as an autotransformer.

You could use a 208-to-32 volt transformer rated 640 VA or so.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

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