Resistance Spot Welder Service

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jmbowles78

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My first post here, so go easy on me. :)

Our company recently received some Japanese Resistance Spot Welders with the following nameplate information:

Rated Capacity: 60KVA
Max Weld Current: 18000A
Max Input: 140KVA
Primary Voltage: 480V
Duty Cycle: 7%

I'm trying to make sure we have a circuit that can power one of these welders at a time (we have 480V 60A fused disconnects). Unfortunately, these are used welders and we don't have any manuals or I would check there first.

Based on NEC 630.31(A)(2), I've done the following calculation:

60KVA*1000 / 480V * 0.27 = 33.75A

Am I missing anything in the calculation, or should we be okay?
 
D

Designer69

Senior Member
the numbers on your calc dont work out but I'm glad you asked this question because I am curious to learn about wiring welders.
 
K8MHZ

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
Realize that the weld current will be very high at very low voltage, typically 1.5 to 5 volts.

The duty cycle is relative to how long the power is going to the tips, usually less than a second, compared to the rest of the cycle, which is complete when the machine opens up and is ready for a new part.

So it's important to realize that for a very short time there will be a very high demand followed by a period of time with relatively low demand and the nameplate values may not provide the exact information you need. Simple (in my head) math looks like you will need hundreds of amps for the weld cycle but with such a short duration. Far smaller of a supply along with inverse time breakers or dual element fuses is what is required.

Oh, and it looks like you multiplied the duty cycle by the amps to get the size. I don't think that's how it's done.
 
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steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
I would start with the 140KVA max. input rating, and use the 480V to get an max input current. (I assume these have a 3 phase input).

Then you might try looking through the NEC to see if there are any demand factors specifically for welders that you could apply. Of hand, I'm not sure if there are any or not.

Steve
 
K8MHZ

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
steve66 said:
I would start with the 140KVA max. input rating, and use the 480V to get an max input current. (I assume these have a 3 phase input).

Then you might try looking through the NEC to see if there are any demand factors specifically for welders that you could apply. Of hand, I'm not sure if there are any or not.

Steve
Click to expand...

we have 480V 60A fused disconnects
Click to expand...

Nowhere near 140,000 VA.
 
J

jmbowles78

Member
To everyone,

Thanks for the feedback. It is much appreciated.

K8MHZ said:
Realize that the weld current will be very high at very low voltage, typically 1.5 to 5 volts.

The duty cycle is relative to how long the power is going to the tips, usually less than a second, compared to the rest of the cycle, which is complete when the machine opens up and is ready for a new part.

So it's important to realize that for a very short time there will be a very high demand followed by a period of time with relatively low demand and the nameplate values may not provide the exact information you need. Simple (in my head) math looks like you will need hundreds of amps for the weld cycle but with such a short duration. Far smaller of a supply along with inverse time breakers or dual element fuses is what is required.

Oh, and it looks like you multiplied the duty cycle by the amps to get the size. I don't think that's how it's done.
Click to expand...

Marky,

This is exactly what I was thinking. I wish I knew what the math was in reference to the Milller chart. Somehow they get from 20KVA @ 40% duty cycle to 460VAC 45A input. It looks like they might have taken 20,000 / 460 = 43.5A and just rounded up (pure conjecture by me).

The NEC codes states essentially that if you know the actual primary current and duty cycle, you can size your conductors by multiplying the actual primary current times the square root of the duty cycle. I just can't figure out how to get the actual primary current other than measuring once it is hooked up. It seems like my extrapolation of KVA rating / Input Voltage may not be a good method for determining maximum actual primary current.

Our building service is rated for 1000A, so it seems like we might have to buy a breaker that can be sized appropriately.
 
J

jmbowles78

Member
One more post (just for my peace of mind) :-?

Assuming rated capacity means tip output capacity for the welder: 60,000VA / 18,000A = 3.33VDC tip voltage (this seems reasonable)

The NEC states that for variable primary currents and duty cycles for resistance welders, conductor ampacity should be calculated as 50% of rated primary current for manually operated welders.

Rated Primary Current = 140,000VA (rated primary power)/ 480VAC (primary voltage) / 1.73 (3-phase constant) = 168.6A
To me this describes the amount of current I need to supply without dropping the voltage significantly while welding.

Conductor ampacity = 168.6A (rated primary current) x 0.5 (duty cycle multiplier) = 84.3A
To me this describes that although the welder will pull 169A while welding, a manual welder will almost certainly have less than a 25% duty cycle (the 0.5 multiplier is equivalent to the square root of 25%), so the wires will only be heated part of the time with the 169A load. Therefore, 85A rated conductors are acceptable.

The NEC also states that if the actual primary current and duty cycle are known and remain unchanged, conductor ampacity shall not be less than the product of the actual primary current and the duty cycle multiplier (shown in NEC table, basically the square root of the duty cycle percentage). I don't have the actual primary current, so I will use the rated (maximum?) primary current for the calculation.

Rated Primary Current = 168.6A (calculated earlier)
Rated Duty Cycle = 7% (from nameplate)
Duty Cycle Multiplier = 0.265 (square root of 7%)
Conductor Ampacity = 168.6A * 0.265 = 44.68A

To sum up. If my calculations are correct (highly debatable, I'm sure), if I want to be safe, I should size my conductors/fuses for >175A and forget about it. If I go with NEC standard recommendations (25% duty cycle assumption), I can size my conductors for 85A and use dual-element fuses that won't blow during the short period of 169A demand. If I assume that my duty cycle shall not go above 7%, I may be able to use conductors sized for 45A, and dual-element fuses.

Can someone smarter than me review these calculations and tell me if I make any sense? :roll:
 
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