Measuring Power and Gas?

admin

I have a kinda crazy question...I work in the automotive industry as a maintenance tech.  I have two robotic mig welders that I program and a bunch of other resistance welders. Yesterday I received an email from my companys VP.  He was asking if I can "measure the amount of power and gas that our robotic mig welder uses in a 1 inch weld".  I know its a off the wall question, but I was wondering if its possible to answer?  The robots use 460v and it says that it has a current draw of 33 amps.  I know P=VI so is my answer 15180w?   thinking about this too much I was hopeing someone could help me elaberate.Thanks again,Chris
Reply:Hi, I am certainly no electrician, but I will take a stab at this. I think that the current draw on your welder would depend on the amperage being output doing a particular weld, so in other words, if you are doing a weld seam on thin steel such as maybe using 100amp and 20V and compare the current that the same welder used doing thick steel at 400amps at 35V then I think that you will find, that if you measured the welders amp input drain then you would get a significant variation in the wattage/hour used by the welder.The gas usage should be a little bit easier to work out. If as an example, you set your gas at 13 lt/minute and if you can work out how many inches of weld (at a given bead size) you lay down in a minute. Divide the inches into the liters and and you should be able to work out how many inches of weld per liter, you then work out how many liters in a tank of gas and from that you can work out you roughly how many inches of weld per tank, find out how much the tank costs and Divide that by the number of weld inches and you should get a rough estimate per bead size. I say rough because there are quite a number of variables. Tanks fills are measured by pressure, which is also dependent on the temperature of the cylinder, how fast it gets filled, etc. so they don't always have the same amount of gas. Also if the tank is hot from sitting out in the sun or a hot environment then the tank pressure increases but the volume out of the nozzle stays the same. If the tank cools down overnight then its internal pressure will drop and the volume of gas contained will be less at the colder temp but you will use more of the gas in the tank even though it is still using 13lt/min out of the nozzle.You would have to do this over a long time and average these measurements out.At least this is what I think happens, I maybe talking out of my backside but someone in this forum who knows more about this may correct me.Lastly the type of welder (inverter or transformer based) will also effect the input drain as Inverter types use considerable less then the transformer types whether they are actually welding or just in idle mode between welds. When I setup my 3 phase power to run my Miller Auto Invision 456 (inverter type) I installed a 3 phase meter to work out my wattage usage for billing purposes. I put though over half  a 15Kg spool through it and I thought that my meter wasn't working properly, I was expecting to sell one of my kidneys to pay for the bill, previously using a transformer based MIG I had a considerable increase in my electricity cost even though the Miller Auto Invision 456  was using many more weld amps and higher voltage by welding on heavier steel. It wasn't a scientific test but was enough for me to not worry about adding any extra in my invoices. The gas and wire cost was much more significant to me.The manufacturer of your welders should have those answers so I would contact them first or better still advise your  VP to ring them up, make him earn his keep, it shouldn't be your responsibility to do this anyway. Just my 2 cents worth. Ed.Miller Auto Invision 456 + S-62 wire feederC6240B1 Gap bed lathe16 ft3 air compressor16 speed pedestal drillHafco BS-912 Bandsaw
Reply:The metering you're looking for is power into the machine from the wall outlet. Ed is the closest to correct, but the challenge is knowing the voltage being supplied by the machine for a given weld from the supply's output. You mentioned that you're supplying the machine with 460V, so what you would really want to know is current from the mains. It gets more complex than just sticking an Ammeter in series with a line because if your machine does not have any power-factor correction, the actual power expended will be a complex quantity (so your reading won't necessarily be the most accurate). Realistically, if this is a big deal your company may want to invest in some industrial supply monitors from a company like Schweitzer Engineering. If it's the back of the welding unit which lists 33A max draw, then you can half-*** a number through duty derating. Maximum power under load would be 460x33= 15.18kW as you calculated. If you know the maximum settings for the welder (say 300A for a TIG) then you turn the effectively consumed power to a percentage. E.g. the given weld under study needs 200A, Instantaneous Power Input=15.18kW * (200A/300A) = 10.12kW. The next component of the cost calculation is how long you're welding. If the 1" length takes 10 seconds to create, you extrapolate it into fractions of an hour: 10s * 1min/60s * 1hr/60min = /3600 = hours of welding time. For that 1 inch at 10 seconds, that's 0.00277 hours. You're billed on the kW per hr, so 10.12kW * 0.00277hrs = 0.028kWhIf you're charged 6c/kWh, that 1 inch costs the company 0.168c.Still remember, this is going to be a very crude calculation because you don't have all the information.