Home |
Search |
Today's Posts |
![]() |
|
Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
|
LinkBack | Thread Tools | Display Modes |
#1
![]() |
|||
|
|||
![]() "Ignoramus18928" wrote in message ... I am confused. The nameplate on my hobart cybertig says that at 230V, it consumes 44 amps. http://igor.chudov.com/projects/Weld...r/dscf0003.jpg Absent the power factor number, that means roughly 220*44*sqrt(3) = 16 kW power consumption. Yet, it is a 28V, 200A welder, which means approximately 6 kW power output. How can I reconcile these numbers? How can it take in 16 kW and yet only produce 6 kW? That seems awfully wasteful. I tried to estimate what it would take, assuming more reasonable power efficiency. It if produces 6 kW and wastes 2 kW as heat, then it would need 8 kW. Estimating amps, 8 kW/220/sqrt(3) = 21 amp. This is not an idle question for me. I have 3 phase cabling that is capable of conducting 21 amps (combo of a 12 and 10 gauge cables). For 44 amps, I would need to spend big bucks to buy 4-6 flexible cable. Also, I can easily support 8 kW power consumption after adding an extra idler to my RPC. 16 kW is out of my realm of possibility due to branch circuit capacity and common sense. Some clarification would be appreciated. I always thought the rated input current was the peak current (ie when you initially strike an arc), with the rated output being the continuos current? |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
Running Lincoln Single Phase 220V welder from generator | Metalworking | |||
3 phase welders | Metalworking | |||
Low priced ac/dc welder on ebay ( single phase ) | Metalworking | |||
Re. Rotary phase converters - magic or myths | Metalworking | |||
static phase converter question | Metalworking |