Help choosing TIG machine

Dear Sirs,
My brother is gonna buy a TIG machine but at this time is very confused in wich one he will decide for. Concerning adjustment parameters they are very similar,so,the main problem is the power consumption Please find some data below.
MACHINE 1 Mains voltage 230 V Welding current range TIG 3 - 220 A Welding current range Electrode 10 - 180 A Duty cycle (10min/40°C) 40% at 220 A
MACHINE 2 Mains voltage 3 x 400 V
Duty cycle 100% 190 A / 27,6 V
Duty cycle 60% 240 A / 29,6 V
Duty cycle 25% 300 A / 32
Current range DC 5 - 300
Is there any easy way to calculate power consumption in both ones ??
Any help will be helpfull
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On Sun, 11 Jul 2004 18:04:09 +0100, vmg wrote:
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Using Ohm's Law will be of help here, but you'd need to know the voltage at the TIG or stick end as well as the current draw. I'm not sure how to do the calculation for the 3 phase machine, but for the single phase machine the formula would be:
Output Watts = Output Volts X Output Amps
Input Amps = Output Watts / Input Volts
Input Watts = Input Volts X Input Amps
For example, assume voltage at the TIG torch of Machine #1 is 50 volts (I have no idea what the actual voltage would be, so this is just to illustrate the use of the formula)
11,000 watts = 50 volts X 220 amps
47.82 amps = 11,000 watts / 230 volts
11,000 watts = 47.82 amps X 230 volts
The reason we come back to exactly 11,000 watts in the last equation is because we're ignoring power loss due to inefficiency in the voltage conversion process, etc., within the welder.
I can't give you percentages of a given machine's efficiency, but I think the general rule of thumb is that "inverter" based machines are more efficient than transformer-based ones.

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in
This Lincoln Electric article on energy efficiency adds to the good information Artemia gave you. http://www.lincolnelectric.com/knowledge/articles/content/energyefficiency.asp

On Tue, 13 Jul 2004 18:43:45 -0500, Lance wrote:
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A "bang for the buck" estimate of the efficiency of a given welder can be extrapolated from its input power requirements and its output power. For example, let's say that a particular machine has a stated input power requirement of 50 amps @ 230 volts from the mains, and its rated welding amperage is 300 amps (output amps) with 35 volts at the arc (output volts).
Input Power (watts) = 11,500 = 50 amps X 230 volts Output Power (watts) = 10,500 = 300 amps X 35 volts Power Loss (watts) = 1000 = Input Power - Output Power
So 1000 watts are "lost" to things such as powering the cooling fan, gas solenoid, (and in the case of a MIG welder, the wire feed motor), and losses due to internal resistances such as in the transformer windings and control circuitry, etc.
To express this loss as an efficiency factor, divide Output Power by Input Power:
Efficiency Factor = 0.91 = Output Power / Input Power
Note that this factor may not agree with advertised numbers because manufacturers may not include the power required to run cooling fans, etc. in the calculation. So the efficiency rating above should be taken as an overall, or real world, efficiency rating.