On 03/29/2013 03:59 PM, 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.


IANAE. However, it sounds like they're talking about the VA rating of
the circuit rather than the transformer itself. 9580 VA = 1000 VA *
230/24 .

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net