Spot welder 'specific operation' criteria under 630.31(A)(2)?

[tortuga]

Greetings all
Does a spot welder meet the 'specific operation' criteria under 630.31(A)(2)?
I am doing a spot welder branch circuit calc as per [2020 NEC] 630.31 and OCPD sizing per 630.32.

The welder will only occasionally be used. Looking over the example duty cycle in the FPN I can't imagine the duty cycle to be more than 2%.

Its for a 20 year old single spot welder 480V, the company that made the welder is no longer in operation in the US, the only website I can find is in Korean:

https://www.chowel.co.kr

The name plate did not give the primary kVA, but it gives the 50% duty cycle as 150 kVA
Model is APR-150A-R12PF
Manufacturer is Americian Chowell


I believe its a single phase 480V connection.
The 'Max Input Power' of 858 kVA seems way off, ignoring that I am extrapolating the kVA by:
Reversing the 50% duty cycle on the nameplate: 1/.71 = 1.4
kVA = 150 X 1.4 = 211

T630.31(A) value is .22 for 5% or less
So 211 X .22 = 46.42 kVA
or minimum supply conductor = 97 Amps at 480V

If I stick to the 50% duty cycle I'd be at 312 Amp conductor size, which seems way off.
If I am wrong and the welder has a 3 phase primary it would be at 175 Amps which is within reason but still seems extreme.

 

[topgone]

IDK how you compute but the welder kVA is not the "effective" kVA that you will be designing for!
The heating of the transformer winding is directly proportional to the nameplate kVA times the square of the Duty Cycle (remember heat loss = I^2 X resistance). That said, your multiplier will be (0.50)^2 = 0.25. In your case, 150 kVA x 0.25 = 37.5 kVA! At 480V, your effective amps will be 78A! Does this figure make sense?
If your duty cycle goes down to say 25%, your "effective current draw" will be (0.25)^2 x 150,000/480 = 20A. I will search for the article on the web which shows how welder capacities are computed!

 

[tortuga]

IDK how you compute but the welder kVA is not the "effective" kVA that you will be designing for!
The heating of the transformer winding is directly proportional to the nameplate kVA times the square of the Duty Cycle (remember heat loss = I^2 X resistance). That said, your multiplier will be (0.50)^2 = 0.25. In your case, 150 kVA x 0.25 = 37.5 kVA! At 480V, your effective amps will be 78A! Does this figure make sense?
If your duty cycle goes down to say 25%, your "effective current draw" will be (0.25)^2 x 150,000/480 = 20A. I will search for the article on the web which shows how welder capacities are computed!

Thanks for the reply, I was not able to determine the welder kVA from the nameplate, the only thing meaningful the nameplate gave me was the 50% rating.
So I used the inverse of a 50% rating to extrapolate the 100% rating of 211 kVA:
1/.71 = 1.40
150X 1.40 = 211


Then I use Table 630.31(A) to Adjust the 100% kVA to 5% duty cycle, multiplier of .22:
210 X .22 =46.42 kVA.
46.42 kVA = 97 Amps @ 480V 1 phase.

630.32(A) not more than 200% OCPD size = 175 A
Does this sound correct?

We're going to gather some more info to get a duty cycle;
the max weld times and the welds per min.
Thanks for your feedback

 

[synchro]

So I used the inverse of a 50% rating to extrapolate the 100% rating of 211 kVA:
1/.71 = 1.40
150X 1.40 = 211

You need to multiply by the √(0.50) ≅ 0.71 factor to get the 100% kVA rating instead of dividing by it. The 100% duty cycle kVA rating is less than the 50% rating because it has more time to get hot.

 

[tortuga]

You need to multiply by the √(0.50) ≅ 0.71 factor to get the 100% kVA rating instead of dividing by it. The 100% duty cycle kVA rating is less than the 50% rating because it has more time to get hot.

OK right duh. I was trying to apply table 630.31(A) to the transformer.
The manufacturer is just up rating the transformer as if its only used 50% the time.
So I dont need to do anything to the 150 kVA (313 amps) rating
I just apply the duty cycle to that 50% rating?
150 kVA * .22 = 33 kVA or 69 amps @480V single phase

 

[synchro]

... I was trying to apply table 630.31(A) to the transformer.
The manufacturer is just up rating the transformer as if its only used 50% the time.
So I dont need to do anything to the 150 kVA (313 amps) rating
I just apply the duty cycle to that 50% rating?
150 kVA * .22 = 33 kVA or 69 amps @480V single phase

You would multiply the 150 kVA rating by √(0.50) ≅ 0.71 to get the continuous rating (i.e., 100% duty cycle rating), and then multiply by 0.22 for the duty cycle in your application. 150 kVA * 0.71 = 106 kVA and therefore the minimum ampacity for conductors is (106 kVA * 0.22) /480V = 48.6 A.

Voltage drop should definitely be considered because of the high peak current that will be drawn, and this could effect the conductor size that should be used depending on how long they will be. Preferably the peak current should be determined by taking the weld current that will be drawn and then reflecting it back to the transformer's primary using the turns ratio.

 

[tortuga]

OK thanks got it now.
synchro and topgone thank your for the enlightenment on how to do this the correct way.
The spot welder was probably used in some form of assembly line and now will be occasionally used in a welding shop.
I am probably going to go with some #3 THHN on a 100A breaker.

 

[Bluegrass Boy]

tortuga, I am going to add some pdf’s on here for you to browse through. Some of it will not pertain to your specific situation, but it will give some extra information to add to your thoughts of what may be necessary for the welder to produce a good weld.
I hope these help.