PE (always learning)
Senior Member
- Location
- Saint Louis
- Occupation
- Professional Engineer
Hey everyone,
Hope you are all doing well. I am doing a project for a high school where they are adding a stand alone building for a shop class. This shop class will have several arc welders with dedicated circuits to each one. I'm a little new to article 630 and just wanted to verify my assumptions are correct. The two types of arc welding units being used on this project are the Miller Thunderbolt 210 and the Millermatic 252. Also, the electrical service for this building is 120/208V, three phase. I have provided fusible disconnects at each welding station that then feed into each welding outlet. The Thunderbolt 210 specifications show a rated primary current of 43.4 amps at 20% duty cycle. So I would use Table 630.11(A) for my calculation for the conductor size, and this would be 43.4 amps x (0.45) = 19.53 minimum conductor ampacity. According to this, I could use a #12 AWG minimum, but I decided to go with a #8 for voltage drop. Next I would size the overcurrent protection for this welding unit using 630.12(A). The breaker can be rated no more than 200% of the welder current which would be 200% x 43.4 amps = 86.8 amps. I went with an 80 amp breaker back at the main panel supplying this welder.
The millermatic 252 shows a rated primary current of 64.2 amps at 40% duty cycle. So I would again use Table 630.11(A) for my calculation for conductor size, and this would be 64.2 amps x (0.63) = 40.5 minimum conductor ampacity. According to this, I could use a #8 AWG minimum, but I decided to go with a #4 for voltage drop. Next I would size the overcurrent protection for this welding unit using 630.12(A) again. The breaker can be rated no more than 200% of the welder current which would be 200% x 64.2 amps = 128.4 amps. I went with a 125 amp breaker back at the main panel suppling this welder.
Welding unit specifications are attached for reference. Please let me know if I'm in interpreting article 630 correctly or if I have made a mistake.
Best Regards,
Engineer in training
Hope you are all doing well. I am doing a project for a high school where they are adding a stand alone building for a shop class. This shop class will have several arc welders with dedicated circuits to each one. I'm a little new to article 630 and just wanted to verify my assumptions are correct. The two types of arc welding units being used on this project are the Miller Thunderbolt 210 and the Millermatic 252. Also, the electrical service for this building is 120/208V, three phase. I have provided fusible disconnects at each welding station that then feed into each welding outlet. The Thunderbolt 210 specifications show a rated primary current of 43.4 amps at 20% duty cycle. So I would use Table 630.11(A) for my calculation for the conductor size, and this would be 43.4 amps x (0.45) = 19.53 minimum conductor ampacity. According to this, I could use a #12 AWG minimum, but I decided to go with a #8 for voltage drop. Next I would size the overcurrent protection for this welding unit using 630.12(A). The breaker can be rated no more than 200% of the welder current which would be 200% x 43.4 amps = 86.8 amps. I went with an 80 amp breaker back at the main panel supplying this welder.
The millermatic 252 shows a rated primary current of 64.2 amps at 40% duty cycle. So I would again use Table 630.11(A) for my calculation for conductor size, and this would be 64.2 amps x (0.63) = 40.5 minimum conductor ampacity. According to this, I could use a #8 AWG minimum, but I decided to go with a #4 for voltage drop. Next I would size the overcurrent protection for this welding unit using 630.12(A) again. The breaker can be rated no more than 200% of the welder current which would be 200% x 64.2 amps = 128.4 amps. I went with a 125 amp breaker back at the main panel suppling this welder.
Welding unit specifications are attached for reference. Please let me know if I'm in interpreting article 630 correctly or if I have made a mistake.
Best Regards,
Engineer in training
