On Wed, 15 Jul 2009 12:57:59 -0700, "Leo Lichtman"
wrote:


"Robert Swinney" wrote: Sorry, I mean to write: Jon it oughta be OK. The
spots will just be 5/6th of normal size.
^^^^^^^^^^^^^^^^^^^
No, I think you need to consider the *area,* not the diameter. The spots
will be 25/36 the normal size. G



I think we should plug in a few numbers.

The operating frequency changes from 60 hz to 50Hz but not the
supply voltage

The transformer ratio and hence the open circuit output
voltage is unchanged.

The copper loss is almost unchanged unless the iron circuit
aproaches saturation.

If the transformer is conservatively designed the working flux
density will still be well below saturation flux density. Iron
losses are non-linear but increase by very roughly as the inverse
square of the frequency change i.e. 60/50 squared about 1.4x
Typically the iron losses are about half the total transformer
losses so this is equivalent to about 20% increase in temperature
rise or, for intermittent use, 20% decrease in duty cycle.

If it's light weight low cost design running as close to
saturation as they dare then there will be a larger increase in
iron loss. The welder will still work but you may have to
severely limit the duty cycle to keep the temperature rise within
limits.

Summing up, your welder should work fine on 50Hz.
It is delivering the same output voltage so the spot size will be
unchanged. The transformer will run hotter so you should be
conservative with your duty cycle.

If you are in any doubt about the 50 Hz temperature rise,
you can play safe by reducing the supply voltage. 5/6 approx 100V
would restore the 60Hz flux density but this gives a 20% drop in
output voltage.

Heat output is proportional to V squared so this is equivalent
to a 44% downsizing of the welder capability with proportionate
effects on spot area and permissible sheet thickness.

Jim