Conductors and OCPD for welder. (630.11(a))

[Dsg319]

Would this be correct in saying this or do I have it wrong. Will post pics of specs from online. But anyhow (non motor)Lincoln 225 arc welder.

It has a rated input current of 50amperes.
Does not list supply conductor size or OCPD. Would it be correct to use 630.11(a).

If that’s the case it would be perfectly fine to supply it with #8awg (60degree) and a 40ampere breaker. Am I doing my math right?

Primary current x multiplier x duty cycle

225 x 45%x 20%= 20.25ampere conductors

 

[suemarkp]

I don't think so. The output current is mostly irrelevant to sizing the circuit. The things that matter are the input current and the duty cycle. The input current is 50A and the duty cycle is 20%. So 50 * .45 = 22.5A. So you could supply this with a #12 circuit on a 100A breaker if you really wanted to. I would not...

8 AWG copper is almost always OK on a 50A receptacle and breaker if the wire and receptacle are 75C rated (most 50A receptacles have a 75C marking). Dropping the breaker to 40A may cause the breaker to trip. I would keep it at 50A, and you are permitted to go up to 200% rated input current if there are tripping issues. The code use to require "Welder use only" labeling on receptacles with undersized wires and/or oversized breakers intended for welders. I can't find it, but it is certainly a good idea if you do that. Even safer is don't undersize anything if there is the possibility of different things utilizing that receptacle (e.g. someone decides to plug a Tesla charger into it).

 

[Dsg319]

I don't think so. The output current is mostly irrelevant to sizing the circuit. The things that matter are the input current and the duty cycle. The input current is 50A and the duty cycle is 20%. So 50 * .45 = 22.5A. So you could supply this with a #12 circuit on a 100A breaker if you really wanted to. I would not...

8 AWG copper is almost always OK on a 50A receptacle and breaker if the wire and receptacle are 75C rated (most 50A receptacles have a 75C marking). Dropping the breaker to 40A may cause the breaker to trip. I would keep it at 50A, and you are permitted to go up to 200% rated input current if there are tripping issues. The code use to require "Welder use only" labeling on receptacles with undersized wires and/or oversized breakers intended for welders. I can't find it, but it is certainly a good idea if you do that. Even safer is don't undersize anything if there is the possibility of different things utilizing that receptacle (e.g. someone decides to plug a Tesla charger into it).

I believe you are correct, I learnt how to do it once in a book but since have forgotten and I forgot to mention it is (non metallic cable) Romex. Which leaves me at 60degree ampacities right.

 

[Dsg319]

StackPath

In the link Is a good example of sizing circuit conductors for a welder (transformer type).

 

[Dsg319]

I don't think so. The output current is mostly irrelevant to sizing the circuit. The things that matter are the input current and the duty cycle. The input current is 50A and the duty cycle is 20%. So 50 * .45 = 22.5A. So you could supply this with a #12 circuit on a 100A breaker if you really wanted to. I would not...

8 AWG copper is almost always OK on a 50A receptacle and breaker if the wire and receptacle are 75C rated (most 50A receptacles have a 75C marking). Dropping the breaker to 40A may cause the breaker to trip. I would keep it at 50A, and you are permitted to go up to 200% rated input current if there are tripping issues. The code use to require "Welder use only" labeling on receptacles with undersized wires and/or oversized breakers intended for welders. I can't find it, but it is certainly a good idea if you do that. Even safer is don't undersize anything if there is the possibility of different things utilizing that receptacle (e.g. someone decides to plug a Tesla charger into it).

It seems to me that code language calls input current, primary current. I’m assuming anyhow.

 

[suemarkp]

Yes. The welder is basically a transformer. The primary is typically 120, 208, 240, or 480 VAC and is the input. The secondary is lower voltage DC at a much higher current and is the output.

 

[retirede]

Yes. The welder is basically a transformer. The primary is typically 120, 208, 240, or 480 VAC and is the input. The secondary is lower voltage DC at a much higher current and is the output.

Not always DC. The aforementioned “Buzz Box” that started this thread is AC only.