# Resistance welders conductor sizing(630.31)(a)(2)

#### Dsg319

##### Member
I can’t quite seem to fully understand this one. Could anyone shed some light for me? Could be I’m just having a blonde moment at this second lol

#### Attachments

• 333.4 KB Views: 34

#### Eddie702

basically the way I understand it if you have a welder with a 50% duty cycle ( you can only weld 30 min out of 1 hour at full welder capacity) the conductors feeding the welder only need to be 71% of the welders nameplate.

Welding machines seldom run at full capacity so when the welding output is low the input is low

• Dsg319

#### topgone

##### Senior Member
The sizing of the conductors for welders is based on the effective current drawn by the unit. By "effective" means the current is derived by multiplying the rated current of the welder unit at 100% duty by the square root of the duty cycle. At a 50% duty cycle, the effective current will be rated current X square root of 0.5 = 0.71!

#### synchro

##### Senior Member
The amount of conductor heating at any given time is proportional to the current squared ( i.e, I²R ).
The effective current that topgone mentioned is the amount of continuous current (i.e., at a 100% duty-cycle) that would produce the same amount of heating as the actual current that has a duty cycle D%.
D% = 50% means the welder is on for 1/2 of the time during the standard 10 minute period.

Therefore to get an equivalent amount of heating:
(effective current)² x 100% = (actual current)² x duty-cycle D%
After dividing by 100% and taking the square root on both sides of the equation:

effective current = actual current x √( duty-cycle D% / 100%)
And so the multiplier to use for calculating effective current is: √( duty-cycle D% / 100%)

And so as topgone mentioned you get the numbers in the table such as:
√( 40% / 100%) = 0.63
√( 25% / 100%) = 0.5
...etc.

#### ozman

##### Member
I like this thread because I’m having trouble understanding the calculations on a question from a psi practice exam. Is there anyone that can help me out?

Multiple nonmotor generator arc welders are being served by a branch panel. There are two 120-amperes welders with a 70% duty cycle, one 90-amperes welder with a 80% duty cycle, and one 250-amperes welder with a 50% duty cycle. What is the minimum disconnect for this branch panel?
i keep getting 459 but the answer it shows is 450. Am I just to round or did I do something wrong?

#### synchro

##### Senior Member
I get 458A, so apparently they are rounding off their answer.

#### Fred B

##### Senior Member
I like this thread because I’m having trouble understanding the calculations on a question from a psi practice exam. Is there anyone that can help me out?

Multiple nonmotor generator arc welders are being served by a branch panel. There are two 120-amperes welders with a 70% duty cycle, one 90-amperes welder with a 80% duty cycle, and one 250-amperes welder with a 50% duty cycle. What is the minimum disconnect for this branch panel?
i keep getting 459 but the answer it shows is 450. Am I just to round or did I do something wrong?
I get 458A, so apparently they are rounding off their answer.
NEC 2017
630.11(B) Grouping of Welders ....currents determined in 630.11(A) as sum of 100 percent of the two largest welders, plus 85 percent of the third largest welder, plus 70 percent of the fourth largest welder ......
You were calculating at 100% for all loads
Thus my calculations I have 250A x .71 multiplier @100% =177.5, plus 120A x .84 m @100% = 100.8 plus, 120A x .84 @ 85%= 85.68 plus, 90A x .89m @70%= 56.01 total sum equals 420.05. Non conforming size allows next size up, per table 240.6 equals breaker size of 450A

#### ozman

##### Member
The amount of conductor heating at any given time is proportional to the current squared ( i.e, I²R ).
The effective current that topgone mentioned is the amount of continuous current (i.e., at a 100% duty-cycle) that would produce the same amount of heating as the actual current that has a duty cycle D%.
D% = 50% means the welder is on for 1/2 of the time during the standard 10 minute period.

Therefore to get an equivalent amount of heating:
(effective current)² x 100% = (actual current)² x duty-cycle D%
After dividing by 100% and taking the square root on both sides of the equation:

effective current = actual current x √( duty-cycle D% / 100%)
And so the multiplier to use for calculating effective current is: √( duty-cycle D% / 100%)

And so as topgone mentioned you get the numbers in the table such as:
√( 40% / 100%) = 0.63
√( 25% / 100%) = 0.5
...etc.
Hey i replied to the
I get 458A, so apparently they are rounding off their answer.
ok but if that’s minimum
NEC 2017
630.11(B) Grouping of Welders ....currents determined in 630.11(A) as sum of 100 percent of the two largest welders, plus 85 percent of the third largest welder, plus 70 percent of the fourth largest welder ......
You were calculating at 100% for all loads
Thus my calculations I have 250A x .71 multiplier @100% =177.5, plus 120A x .84 m @100% = 100.8 plus, 120A x .84 @ 85%= 85.68 plus, 90A x .89m @70%= 56.01 total sum equals 420.05. Non conforming size allows next size up, per table 240.6 equals breaker size of 450A
OH I see what you did. I didn’t account the second 100% to the equation . Thank you!
so much I might have more questions later. I love this community of threads so helpful.

#### oldsparky52

##### Senior Member
• 