computing welding power consumption

One other related question. In the welding shop at the local community college, the stick welding booths are supplied by an eight pack of Miller welders in a back room. To control the welding current, each booth has a control box (a rheostat I assume) before the electrode holder. Does this mean that the power consumption for any individual welder always stays the same while welding regardless of the setting of the control box in that booth?

Thanks, Stan

Stan

The input current waveform will also be something significantly different than a sine wave. The way your get constant current out of a transformer is by saturating the magnetic field in the core, which means that you will draw a lot of current duing the parts of the waveform when the core is not saturated and a lot less current when you hit the level at which the core saturates. As a result, the input current will have a ton of distortion, and even a clamp-on current meter may not read correctly. You'd need a true-RMS ammeter, such as you would find on a power quality analyzer, to measure actual current.

That's also why welders put a bunch of interference back onto the incoming power line.

In addition, it would be really hard to determine the actual energy dissipated in the arc.

If you need to size wiring, use the current rating, not the power rating. The same would apply if your're trying to run it off a generator. Most consumer type generators are current limited, not power limited, and for something with a terrible power factor like a welder, you'll hit the generator current limit long before you hit the maximum power.

If you give us a little more info as to why you want to know this, we can maybe give you a little more help.

Keith

Are you trying to figure out the energy consumption in watt-hours? That's what you need to know if you're interested in how much it costs to operate your welder.

Do you want to know the volt-amps required by the welder? That's important to be able to size the wiring, circuit breaker and transformer that supply the welder.

Or do you want to know the actual instantaneous power consumed by the welder while it's in operation? That is not so easy to measure. As you surmise, it is not steady, it varies constantly and to measure it accurately you would need some fairly exotic equipment, including a current shunt and probably a storage oscilloscope.

Really, I think I would try to measure the average power consumption using the power meter for your electrical service. Just turn off everything connected to the meter except for your welder. Run the welder without striking an arc for a timed period of several minutes; use the power consumed (in kW-hr) divided by the time (in hours) to figure the average idle power consumption of the welder (in kilowatts). Then you can draw some beads at whatever setting you like, repeating the process to figure out the average power consumption at that particular setting.

In answer to your second question, I'm fairly certain that the power consumption for each welder station will vary with its rheostat setting: the higher the amperage setting, the higher the power consumption.

As far as computations based on the secondary, one has to know the coefficient of mutual inductance and efficiencies and heat and other parasitic losses to transfer that to power consumed in the primary and Im not smart enough to do all that stuff like I might?? have been abole to when an EE student at Purdue in the late sixties. I look at it as volts (say 220) times amps in the primary and the priamry amps depends on the secondary load but again, I just cant say how to compute primary amps versus the secondary load.

Where I used to work the utility penalized and charged us big time for a less then unity power factor. Our friend the good Buick n Deere man can explain all that stuff better then me as he works in the field while I have been retired from it a good while.

Fun discussion but sorry I cant provide you an answer.

John T

Okay, but the problem remains that I don't know the amperage on the primary side or the voltage on the secondary side. Don't both of those change radically the instant the arc is struck? Are those figures available to me only by actual measurement? Also, how do I determine the phase angle? Is it a constant depending on the nature of the power supplied by the utility company (one phase vs. three phase) or is it a value that changes according to factors involved in the work being done?

Do you have any reason in the world to think that I know something about any of this?

Thanks, Stan

Boy you should have asked thease questions 25yrs ago when we were in our prime and actualy using some of this info :)

I can give you some of what I remember but it's up to you to verifiy the info being correct.

You are correct in that a rheostat controls the arc voltage on most of the newer amperage controls.

The secondary side of a transformer power consumption is controlled partially arc voltage, the arc length, as well being influanced by the weld process being used, although most weld processes will be close in both arc voltage and arc amperage in most cases for figuring general math.

As the arc length varies during the welding process so does the potenial voltage vary thus changing the applied amperage or heat to the molten puddle. It's very difficult to make a decent weld if the arc length continously changes. This is one why a machine weldment has such a excellant apparence as arc length is precisely controlled.

Reading thru my past articles you'll notice I always was critical of holding a exremely steady arc length and stressed it's importance. Well now you will know why.

Secondary Examples: Using 7018

18-22v = 70-120amps
22-24v = 100-150amps

With the above info then we can assume secondary power cost to be with a 50% efficiency factor:

Power Cost per inch = (amps x voltage x power cost p/kw) / (deposition x travel speed) / 1000

Power cost is cents per kw travel speed is in in/min

Maybe someone will correct me if I wrong.

T_Bone

About all that can be said is that if a particular breaker never trips while supplying a welder, the circuit probably had adequate capacity. reputable welder makers will specify the circuit size needed. Then that will be only be absolutely needed for running the welder with the largest rods and maximum current. And unless you are in the habit of welding 1/2" slabs with 1/4" rods in one pass, you may never load the welder that hard.

You can learn power consumption with the utility watthour meter, but not current, because the power factor of the typical transformer type welder will be on the poor side causing it draw lagging current likely twice what the watthour meter would indicate. You can measure watts with the watthour meter by finding a value for kh on the face. Often 7.2 watthours per revolution. Time a revolution in seconds. Then that time divided into 3600 x 7.2 is the watts consumed by that load. W = 7.2 * 3600 / t. The load has to be constant for that to be reliable. A welder load often isn't constant.

Then if you run a welder that needs a 30 amp circuit on a 20 amp circuit, it may not weld as well because the overloadedd 20 amp circuit will have more voltage drop which leads to poor current regulation.

Open circuit voltage may vary from 40 to 75 volts, closed circuit voltage will vary with the arc length. Any measurements of primary current will vary with the length of the arc, the size of the rod, and the skill of the operator.

Power factor is determined by the load, the utility prefers a very high power factor. A transformer welder will not present a high power factor under any operating conditions.

You could weld with a plain transformer, but then arc current would vary inversly with arc gap and you would blow holes every time you slipped and shortened the gap. My dad built a DC welder using an aircraft generator and it had that characteristic.

What you want to know is difficult to know and harder to repeat the test.

So what is it you are after? Size of wires to feed the welder? energy consumption of welding?

Gerald J.

I answered the question of what I'm after in a posting to everyone, above. What I'd like to ask you specifically is whether your dad was able to solve that problem with the DC welder made from an aircraft generator. The reason I'm interested is because last year on eBay I bought what was sold as a 200 amp motor/engine driven DC welder, which turned out to be, you guessed it, an aircraft generator. I didn't have anything to run it at the time, or any pressing need to get it running, but I figured that eventually I'd come across a good deal on a 10hp (M/L) engine. A few weeks ago I got an amazing deal on an 11.5hp B&S engine (which happens to be mounted on a completely functional pressure washer) so I'm suddenly much closer to putting together a portable engine driven welder. Except maybe I won't if what I'll be doing is giving up a good 3,500 psi pressure washer to get a not-so-good welder. I'll be most grateful for any information you can give me (also, thanks for the info you've already provided on my initial question.)

All the best, Stan

He wound a large toroid core from fence wire. Probably 10" OD, 5" ID, maybe larger and wound a layer of about #2 aluminum wire through the toroid to give an inductance for better arc control. But the machine worked best while welding 1/4" stock or thicker, it didn't turn down well. Welding 1" water pipe, it often blew holes. After at least a decade of using it, my dad picked up a 105 amp Lincoln at a state fair price was forever pleased. It turned down, it didn't need water, fuel, and battery, and all that space. And it wasn't any less portable. Besides he was able to trade off the model B engine to his brother who wore it out running his saw mill.

Because of the quill shaft, you have to use an outboard bearing with a belt drive. Its not stiff enough to support any side thrust and I suspect it won't take that belt thrust long with a simple bearing out beyond the pulley. It probably won't like anything but a perfect stiff coupling. It won't like power once every other revolution that your 11 hp engine will give and that 11 hp engine probably won't hold the generator load.

The aircraft generator is compound wound with commutating interpoles. Its made to be easily regulated for voltage and the series field windings are connected to improve its voltage regulation which makes its current vary wildly with load and is exactly the reverse of what is needed for good welding. I've not looked inside to see if its possible to easily remove the series windings from the circuit, though the commutating interpole probably help brush life, if the main series field windings had reversed polarity it would be a better current source than voltage source. If you had only the series windings in use with a really big shunt resistor to vary the fieldd excitation and maybe just a tiny big of shunt field excition to get it to have an open circuit voltage, it might make a good welder, but as it is, its rotten, good only for heavy work in most hands.

Those same counpound windings are what make the aircraft generator second rate as a motor for an electric car without using a fancy solid state chopper that wasn't possible 20 or 30 years ago.

I have the little Lincoln, and the aircraft generator, but I'm no good at welding, and I'm unwilling to become a slave to a truck load of batteries. So it may never get used by me.

Gerald J.

That's not good news but I'm sure it will save me loads of eventual frustration. There are a number of things that I'm good at but working with electronics and electrical equipment beyond an elementary level is not one of them.

Contact me if there's ever anything I can help you with. I was a residential remodeling contractor for years (working mostly alone) so I know much of what there is to know about houses, although you can get plenty of good (maybe better) advice about specifics from specialists on this forum. I don't imagine I'm among the top five most knowledgeable welders on this forum, and maybe not the top ten, but I have a lot of the kind of knowledge that you get from approaching it academically, as well as hands on, so maybe I can be of help to you there. I'm also a CPA, although I don't like it much and don't practice beyond doing taxes for and offering advice to friends and family. Nevertheless, I know a lot and I have resources for finding out what I don't already know that probably wouldn't be available or easy for the average non-CPA to understand. So that's another area where I might be able to provide useful advice.

You can e-mail me anytime through this forum or directly at k a t s t a n @ z h o n k a . n e t (just take out all the spaces.)

Thanks again, Stan Lewis