knowingthis
Member
I just want to make sure I'm getting this right. We're just going to relocate this welder, but we'll be running power from a different panel now. I could just use the exact same setup for the existing location but this looks like a good opportunity to get some practice in as I've never dealt with welders. Supply voltage is 480v 1ph.
So for my supply conductors I go to 630.11(A) which tells me I can use the rated primary current of the welder (48A@480v) multiplied by the value given in Table 630.11(A). Since this unit has multiple duty cycles, I'll select the highest, 100%. (48)(1)=48 Table 310.15(B)(16) 60deg. column says I need #6. TANGENT But if it only listed a 40% duty cycle it would be (40)(.63)=25.2 which would drop me down to a #10.
For the OCPD I'll use 200% the rated primary current as per 630.12(A). (48)(2)=96A=100A OCPD. And the circuit breaker in the panel feeding this load can be used as a disconnecting means for this welder as long as it is properly labeled as such.
Say I have three of these welders. 630.11(B) (48)(2)+(48)(.85)=136.8...137=1/0 630.12(B)=300A OCPD
Take that to a gutter and feed three discos fused at 100A (now using the calculation from above for each individual welder) each and drive on down the road. Did I screw something up here? It seems so straightforward I feel like I'm getting something wrong. Lack of familiarity= lack of confidence.
edited to remove massive photo of welder and replace with not so massive nameplate photo.
So for my supply conductors I go to 630.11(A) which tells me I can use the rated primary current of the welder (48A@480v) multiplied by the value given in Table 630.11(A). Since this unit has multiple duty cycles, I'll select the highest, 100%. (48)(1)=48 Table 310.15(B)(16) 60deg. column says I need #6. TANGENT But if it only listed a 40% duty cycle it would be (40)(.63)=25.2 which would drop me down to a #10.
For the OCPD I'll use 200% the rated primary current as per 630.12(A). (48)(2)=96A=100A OCPD. And the circuit breaker in the panel feeding this load can be used as a disconnecting means for this welder as long as it is properly labeled as such.
Say I have three of these welders. 630.11(B) (48)(2)+(48)(.85)=136.8...137=1/0 630.12(B)=300A OCPD
Take that to a gutter and feed three discos fused at 100A (now using the calculation from above for each individual welder) each and drive on down the road. Did I screw something up here? It seems so straightforward I feel like I'm getting something wrong. Lack of familiarity= lack of confidence.
edited to remove massive photo of welder and replace with not so massive nameplate photo.
