Power and Gas usage

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I have an off the wall question... I work as a maintenance tech for an automotive manufacture and I program 2 robotic welders as well as a bunch of other resistance welders. Today I was asked by my companys VP if I can " measure the amount of power and gas used during a 1 inch weld for our robotic welders".  Now I know that P=VI and our robots use 460v and it says 33 amps so is my answer 15180watts?  Now is that the same during the weld process? Im thinking about this too much was hopeing someone can help elaborate.  Thanks again,Chris
Reply:W=VA (Watts equals volts times amps) Is that 33A documented or observed? Ideally you should measure the actual current by using a clamp on ammeter.  Bear in mind that the current will vary based on what the machine is doing.  Clamp on meters can be had for under $100 at Lowes - I use one to take power measurements off our large servers in the data center.Hobart LX235Victor 250 Oxy-Acetylene Rig (welding and cutting)Bobcat 773F-350, 1999, 4x4, 16' 10K# trailerOutdoor Wood Burner - 10 cords/year
Reply:+1 to gwiley on the amprobe. You also need IPM of torch travel to make a calculation.Also if this is for long term calculations for cost over time, you should know the power factor of the equipment in use. It may be buried in the equipment docs or you may need an electrician to determine it, but over a long period of time the meter can show much higher usage than these calculations would indicate if the equipment has a leading or lagging power factor.Matt
Reply:Assuming 460 V, 3 phase your power is 26.3 kva, actual kw will depend on your power factor."The reason we are here is that we are not all there"SA 200Idealarc TM 300 300MM 200MM 25130a SpoolgunPrecision Tig 375Invertec V350 ProSC-32 CS 12 Wire FeederOxweld/Purox O/AArcAirHypertherm Powermax 85LN25
Reply:Norite is spot on... There is power consumption, which is what the power company bills you for.  This is not just Volts * Amps * Hours.  The key point that is missing is how efficiently does that welder use the power.  That is the "Power Factor".A quote from sci.engr.joining.welding:"As for as the power company. They only charge you for the power you use. The power company only charges for power that goes into your shop/home and stays. Power that circulates in and back out from an inductive load, is not charged for. That is because the power company still has it at the end of the day. "That circulation of power in / out of the shop is a function of the power factor.Another quote from sci.engr.joining.welding (Leo Lichtman)"For AC, the equation is P = VA Cos Thetawhere Theta is the phase angle between volts and amps.  Cos theta is the power factor.  For DC, and for non-inductive AC it is 1 (one.) For AC, as the load becomes more inductive or capacitive, the phase angle changes, and Cos theta becomes less than 1. "Additionally, if the welder is transformer based, there will be losses in the transformer windings, which will also consume power.  Odds are, if they want to know kW/inch of weld, you will need to also include the cost of the machine at idle...such as waiting for the next part to move down the conveyor line/weld positioning/weld prep time._kevin