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Absorbed Power

bczygan

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Nov 4, 2009
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DETROIT! Arsenal of Scrappers
I have a 230VAC 1PH welder with a listing of 5.4 kilowatts of absorbed power.

How does this relate to the required circuit size and breaker in amps, and wiring, both cord and circuit portions?

Bill
 
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wyliesdiesels

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Aug 14, 2012
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Modesto, CA
What model welder?

What does the manufacturer call for?

What kind of cord and plug does the welder have?

Welder circuits are wired based on duty cycle and type of welder (transformer or motor driven)....

Without any of that info we would just be guessing...
 
OP
B

bczygan

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Joined
Nov 4, 2009
Messages
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Location
DETROIT! Arsenal of Scrappers
What model welder?

What does the manufacturer call for?

What kind of cord and plug does the welder have?

Welder circuits are wired based on duty cycle and type of welder (transformer or motor driven)....

Without any of that info we would just be guessing...

Doesn't call for anything.

#14, no plug.

HF #44568:

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CCYQFjABahUKEwjj-uPU_-_GAhWBHR4KHbf0AG4&url=http%3A%2F%2Fpdfstream.manualsonline.com%2F5%2F52e92f38-6fd7-43cc-9fb2-92937e3fd196.pdf&ei=YjWwVePBKYG7eLfpg_AG&usg=AFQjCNH9vDv6Uasxabb6WASVSaHXC_Kg2w&sig2=xlqKXF6MRj7td3oq8kSZDQ

Transformer type, I think.
AC/DC
10%@130AC 15%@100DC

Bill
 
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Alchymist

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Central PA
To be "technically correct" absorbed power is true power in a reactive circuit as opposed to reactive power. Power factor is calculated thus:

PF = real power divided by apparent power.
 

sberry

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Jun 18, 2005
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Brethren, Michigan
If it has a 14 wire the breaker needs to be limited to 30A. Use a 12 or better for the circuit.
As a general rule for estimation. The supply wire should be a size larger than the cord on the appliance, there are probably some exceptions but its hard to burn the place down that way.
 
Last edited:

DSMR

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May 24, 2015
Messages
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To be "technically correct" absorbed power is true power in a reactive circuit as opposed to reactive power. Power factor is calculated thus:

PF = real power divided by apparent power.

=cos(theta) :)
 
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