Help choosing TIG machine

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)

47.82 amps = 11,000 watts / 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.

This Lincoln Electric article on energy efficiency adds to the good information Artemia gave you.

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.