I'm hoping one or more of you out there will be able to help out with a couple of questions I have.
Here is some background information about my situation. I just purchased a Hobart Handler 187 wire feed welder. I need to set up a circuit on which the welder can operate. Based on what I have read in the owner's manual, and what I have read on the internet I have come up with the following information.
This Hobart welder has an input amperage of 20.5A and a duty cycle of 30%. Based upon NEC 630-11(a) I can take that 20.5A and multiply it by .55.
So, 20.5A x .55 = 11.275A. Am I correct in interpreting this information to mean that I can use #14 or heavier wire for my circuit? I think this is correct (plus the owner's manual from Hobart says #14 is the minimum wire size allowable).
A second point is that 630-12(a) leads me to believe that in order to determine the correct size of circuit breaker I must not exceed 200% of "the rated primary current of the welder". Does that mean that I take the 20.5A and simply multiply it by 2?
If so, that would give 20.5A x 2 = 41A. I am interpreting this to mean that I cannot use a breaker that is larger that 41A for the circuit (assuming that I match that breaker with a suitable wire size).
At this point I am of the belief that I could run a suitable circuit using #10 wire and a 30A breaker using a 6-50 receptacle.
However, why could I not just wire up a circuit using #6 wire, a 50A breaker, and a 6-50 receptacle? In the future I may purchase a stick welder with 50A input. Could I use such a circuit so that I could plug either machine into the same receptacle? I think this would be less of a hassle than running two separate circuits. I was prepared to install such a circuit but then ran into 630-12(a) on several discussion forums.
Any thoughts, inputs, corrections would be greatly appreciated.