Resistance welders conductor sizing(630.31)(a)(2)

Eddie702

Licensed Electrician
Location
Western Massachusetts
Occupation
Electrician
basically the way I understand it if you have a welder with a 50% duty cycle ( you can only weld 30 min out of 1 hour at full welder capacity) the conductors feeding the welder only need to be 71% of the welders nameplate.

Welding machines seldom run at full capacity so when the welding output is low the input is low
 

topgone

Senior Member
The sizing of the conductors for welders is based on the effective current drawn by the unit. By "effective" means the current is derived by multiplying the rated current of the welder unit at 100% duty by the square root of the duty cycle. At a 50% duty cycle, the effective current will be rated current X square root of 0.5 = 0.71!
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
The amount of conductor heating at any given time is proportional to the current squared ( i.e, I²R ).
The effective current that topgone mentioned is the amount of continuous current (i.e., at a 100% duty-cycle) that would produce the same amount of heating as the actual current that has a duty cycle D%.
D% = 50% means the welder is on for 1/2 of the time during the standard 10 minute period.

Therefore to get an equivalent amount of heating:
(effective current)² x 100% = (actual current)² x duty-cycle D%
After dividing by 100% and taking the square root on both sides of the equation:

effective current = actual current x √( duty-cycle D% / 100%)
And so the multiplier to use for calculating effective current is: √( duty-cycle D% / 100%)

And so as topgone mentioned you get the numbers in the table such as:
√( 40% / 100%) = 0.63
√( 25% / 100%) = 0.5
...etc.
 
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