Rectifier - near welder or near work?

Hi everyone.
Question - cut DINSE couplings into welding leads near transformer or
near work - for coupling and uncoupling rectifier.
Have big oil-cooled welder which stays in hull of a barge (other
barges-in-construction moored alongside).
Thought of these
Near transformer - often less distance to carry the rectifier from
the wharf
Near working end of leads - uncouple rectifier for heavy AC welding
and you'd get at least 20% more power, without having to go back to
the transformer
What does anyone reckon?
Thanks in advance
Richard Smith
Reply to
Richard Smith
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Rectifier near work means lower losses.
Reply to
Phil
Nope. Rectifier near the transformer means that the lines will see lower current after rectifier losses.
Phil wrote:
Reply to
RoyJ
Rectifier loses voltage.
I vote for it does not matter.
i
Reply to
Ignoramus23170
in theory your welder is being fed by something like 575-600 volts 3 phase? maybe 440?
the power loss is much less on the feed side before the rectifier than after because there is less current passing through those lines (At a higher voltage)
IF you are looking to minimize transmission losses and have as much power as possible at the end of the torch then the shorter the Welding side of the leads as possible.
Or am i misreading this? and you have a big AC welder in the hull of the barge and after the AC welding transformer steps the power down you rectify it to rin DC via a rectifier and a bunch of Coke bottle sized capacitors to provide some regulation?
if its the misreading scenario then it really does not matter
Reply to
Brent
Brent -- you read the situation correctly that 440V 3-phase leads feed the big transformer and a rectifier is used to make DC which 6010 and 7018 electrodes can run on. I know there is a lot in the case of the "Rectifier", looking through the air-vents - not electrician so can't work-out what. RS.
Reply to
Richard Smith
Roy - I think Iggy sets us right. Power loss P=I^2R (current * current * the resistance of the leads).
Current is conserved - as electrons are matter, they are conserved and you keep the same quantity of electrons moving in unit time (current). What you've lost in the rectifier is electrical pressure (voltage). Power loss in leads is dependent on the current and not on the voltage.
What I have discovered by experiment, in line with standard guidance, since my original post:
If the rectifier is next to you, you can change between electrode-positive (DCEP) for deep fusion or electrode-negative (DCEN) to get rapid fill with little heat for filling an overly-wide gap joint. I find the difference very useful. So that finally dictates the balance - the rectifier goes at the working end of the leads, so I can swap between DCEP, DCEN and AC.
Thanks Roy and everyone.
RS
Reply to
Richard Smith
Well, duh. You will always get less loss by running higher voltage and less amperage. Many of us run rectifier or diode packs on the AC OUTPUT lines of the welding transformer which is what your question strongly implied.
Richard Smith wrote:
Reply to
RoyJ

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