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irish
January 24th, 2006, 04:07 AM
Ok, I know questions about welders have been asked several times on this forum. I have a question/comment about a Lincoln AC-225 arc welder. According to the current owners manual the welder should be wired to a NEMA 6-50 plug with a 240V (2 pole) 50 amp breaker using a 10 gauge wire, or an 8 gauge wire if the distance of the wire exceeds 100 feet. Can someone please explain why you wouldn't use a 6 gauge wire? Am I missing something here? I thought a 2 pole 50 amp breaker literally allowed 120 volts and 50 amps on each side (pole) not 50 amps total (25amps each side) doesn't this breaker exceed the amp capacity of the wire? Thanks in advance.

Joe

ggratecc
January 24th, 2006, 10:54 AM
First of all, a 50A double pole breaker is 50A each leg (forget about the 25A).
Also, there is no neutral (grounded conductor) in the circuit.

Basically, this is a dedicated circuit for your welder.
Now, I'm not an expert in arc welders or Article 630, however,
NEC 630, B is dedicated to arc welders and cover supply conductor
requirements.
This article is new to me, but since I have my Handbook available...
it states:
Conductors that supply one or more welders shall be protected by an
overcurrent device rated or set at not more than 200 percent of
the conductor rating."

Also states the rating plate shall provide the following info:
....primary voltage, rated primary current, maximum open circuit current,
rated secondary current, basis of rating, such as the duty cycle.

HTH

Roger
January 24th, 2006, 01:19 PM
Sure.....its all about duty cycle. Arc welders are allowed to have their branch circuit conductors derated by a factor based on the duty cycle of the welder. So lets take your AC225 stick welder by lincoln electric. If the nameplate states the I-eff you are required to use it as your minimum circuit conductor ampacity. I've never seen it stated on a stick welder. If that isnt stated you use the rated primary input amps (not output) of the welder. You then take the mutipler factor in the table listed in NEC table 630.11(A) that corresponds to the duty cycle times the primary input current of your welder.

Example: Lincoln AC225 with an input rating of 50 amps to provide the maximum output currents.
The AC225 has a duty cycle of 20% (lincoln website spec for that welder) looking at table 630.11(A) the factor for non-motor welder is .45....

So .45 x 50 = 22.5 amps so this would require 30 amp 10 awg copper wire.... next size up from #12 awg copper which is 20 amp rated.

Now overcurrent protection can be as much as 200% over that rating, so the next overcurrent size up from 22.5 is 25 amp breaker. 200% of that is..........2 x 25 = 50 amps.

So you could breaker at a maximum of 50 amps and use 10 awg minimum for your conductors. You could also breaker at 30 or 40....up to 50.

So the circuit for your welder would be....

Minimum 10 awg copper rated 30 amps.
Maximum OCPD 50 amps.

The #8 awg is allowing for voltage drop.

Why the small wire..... because the welder has a 20% duty cycle. Which means you cannot weld longer than 2 minutes continuous out of a 10 minute time period. Therefore the wire will not get hot enough to damage the insulation. And most welders today have overload circuit protection that normally turns on a fan when they start getting too warm and will trip the internal overload device if they exceed the duty cycle by too much of a margin and shut the welder down.

mdshunk
January 24th, 2006, 07:24 PM
Super reply, Roger.

That's about the most simple and straightforward explanation about welder circuits that I've heard to date.

I generally wire my welder circuits full size, even though the code is more leniant. This allows for an equipment change down the line.

Roger
January 24th, 2006, 08:44 PM
Thanks Marc, it is very hard to explain what the NEC tries their best to disguise. When you read 630.12(a) you have to be careful because it is very easy to misunderstand the intent for the overcurrent protection for the welder....not the supply conductors but the welder. If you arent careful you will think 630.12(a) is saying that the supply conductors to an individual welder can be protected at 200% of the rated primary input current.....or 100 amps.

This confusion is set up in the first paragraph of 630.12(A). In my opinion the first paragraph is speaking to the manufacturer and in terms of groups of welders, in that, if the welder has remote overcurrent protection that protection can be as much as 100 amps in the case of an arc welder with 50 amps rated input.

Then it addresses the case of a welder that does not have seperate protection in the last two paragraphs and the requirements for the supply conductors in that case.....If the supply conductors are protected by an overcurrent device rated at 200% of listed nameplate I-max....Or.....at the rated primary current. Then the welder does not need overcurrent protection.

What a mess 630.12(a) is in its intent.

If you read 630.12(a) its really redundant in the last two paragraphs because it says the exact same thing....twice.

Yep, all the stick welders I've been around have been #6 or #8 awg copper on 50 or 40 amp doubles respectively. I've wired a few for the job and thats the way I did them.

ggratecc
January 25th, 2006, 09:07 AM
Thanks Marc, it is very hard to explain what the NEC tries their best to disguise. When you read 630.12(a) you have to be careful because it is very easy to misunderstand the intent for the overcurrent protection for the welder....not the supply conductors but the welder(internal). If you arent careful you will think 630.12(a) and (b) is saying that the supply conductors can be protected at 200% of the rated primary input current.....or 100 amps.

What a mess 630.12(a) is in its intent.

If you read 630.12(a) its really redundant in the last two paragraphs because it says the exact same thing....twice.


Thanks Roger and Marc bow_1 , I still consider myself a "shipboard" electrician trying to learn the NEC and residential. This forum encourages me to look up
the articles in the NEC, only problem is...I am still misinterpreting them.

Roger
January 25th, 2006, 11:40 AM
ggratecc....Your not alone even the experts cannot agree on 630.12 (a).. ie.. Just go to the Mike Holt forum and see for yourself. I'm sure there are those that will disagree with what I have said.

Also note that I went back and edited out the word internal as it makes it sound like I mean inside the welder as in a overload device.

This sub-section of artcle 630 specifically 630.12(a) addresses "groups" of welders where supply conductors become "feeders" . So dont confuse this with individual welders. Notice the sub-section begins with Welders plural.


A typical design for several welders would be a feeder coming off a buss type supply where it would have ocpd in place for the feeder to the welders. Then at each welder the feeder would be tapped then those tap conductors would be run to a fused disconnect. This solves the NEC requirement for a disconnect for that welder and ocpd for that welder.
Reason is....you dont want a fault with one welder to bring down the rest of them by taking out the feeder ocpd.

So you need to watch out for when it is talking about more than one welder or one welder. I was simply trying to differentiate how you apply ocpd to an individual welder vs. a group of welders. Its very hard to see whats going on in 630.12(A).

I'm still trying to figure out how to explain it better.

In a nutshell...I guess if you are dealing with an individual arc welder you look at the manufacturer specifications and abide by them. There is nothing wrong though in increasing the wire size if you want to address future possibilities.

If you are dealing with an individual arc welder... and you need to run the circuit without the manufacturers recommendations. Then ignore 630.12(A). You stay within 630.11(A) and 630.12 (B) You do not size the OCPD to a individual welder at 200% of the rated input amperage.
Its maximum 200% of the calculated conductor ampacity after you apply the appropriate factor to the rated input current in table 630.11(A).

One last comment... the sizing of the ocpd sorta reminds you of motors where can have a bigger ocpd than the wire size to compensate for in-rush current, that is also what is going on in the case of welders along with duty cycle.

Hope this makes sense... its tough to explain the intent in 630.12(A)