Lincoln TIG 355 input power question

[Nobody]

I recently acquired a Lincoln TIG 355 for use around the home shop, obviously the maximum power requirements exceed what have on tap here but I figured like most of my other stuff if I don't crank it the whole way up it won't be a problem... and thus far it hasn't been for my other equipment. The manufacturer rates the machine at 100A@230/60 at 340A output.

Last night I was messing around melting holes in mild steel (machine was set to 150A but I using a lot less judging from pedal position) with no problems, I put everything down (left the welder on) and ran in the house for a few minutes when I came out the machine was off and the breaker was tripped.

Along with the breaker being tripped it was also warm to the touch as was the plug going into the wall. (Square D HOM breaker with a 50A plug)

Today I borrowed a amp meter (clamp style) from one of the electricians at work, with the TIG welder turned on the meter showed 65 amps of draw (on a 50amp circuit) with nothing else happening, for curiosity sake I had a buddy watch the meter while I struck up an arc... this is where it gets weird. (For me at least)

As I struck up an arc the current into the machine began to drop... and kept dropping in proportion to how far down I had the pedal until with the pedal the whole way down (150A on the machine) the input current was only 37 amps at the breaker. Since I was able to weld for a longer period of time than I was away from the machine without throwing the breaker I have little reason to doubt what the meter is showing.

So now for the questions... why is this thing pulling 65 amps no load? Why does it pull less power from the grid as the power to the work increases?

The machine draws pretty the same current in DC or AC mode TIG or Stick, all of the options were turned off during the measurements except for Hi-freq start.

I inquired with Lincoln on the no-load current rating for this machine but considering it's age and the fact that the Lincoln employees received 18k bonuses this year I probably won't get an answer until about June. :)

[Sandy]

So now for the questions... why is this thing pulling 65 amps no load? Why does it pull less power from the grid as the power to the work increases?

The input to this machine is 230 volt, single phase and should be single phase--------right??? Or is that not correct?? I don't know the machine myself.

Anyway I'll risk going over what you already know---when you measured the current you were clamping around ONE of the two hot input leads?? Bottom line, if you aren't in a weld mode and the machine is in the idle mode there should be very little current draw on the input (house AC) side. Maybe the fan and a tiny loss across any transformers, but minimal.

Either there is a serious fault somewhere or someone has added that pedal and wired it wrong. No work being done (welding current), no current input needed.

[Nobody]

No work being done (welding current), no current input needed.

That has been my experience given my very limited knowledge of electricity. :)

The 355 TIG is an older (early 90s) non-inverter (I believe) machine rated upto 350amps at 40%. I picked it up from a shop that was running the unit on 460 single phase and I tapped it to 230V per the operators manual... a fairly simple procedure.

To answer your questions... yes this is 230v single phase and yes I was going across just one of the hot leads with the clamp meter.

The pedal is a brand new Lincoln unit, even so with the pedal unhooked and the machine running in stick mode it still pulls ~60 amps no load.

Today for kicks I walked out to the shop and turned it on without doing any welding, 5 minutes later the breaker popped.

This evening I took it apart and went over a few things with the wiring diagrams - basically looking for anything that could take that much current without burning up, one I thing noticed are several huge resistors called the loading, stabilizer and background resistors per the wiring diagram. Aside from those and the main transformer I couldn't find anything that could take that much current without instantly vaporizing into a cloud of smoke.

I suppose something could be amiss with the main transformer but I wouldn't expect it to weld as good as it does. Tomorrow I'll go over it with the IR thermo and see what gets hot when it's running with no load.

[Nobody]

Looks like Power Factor Correction might be what is causing this issue.

The TIG 375 (newer) manual shows an idle current draw of 62 amps with PFC.

In the Miller 250DX manual I found a chart that shows that machine pulling 45 amps at idle with PFC but only 3 without PFC. At 60% duty cycle the PFC equiped machine pulled 61 amps to the 82 amps of the non PFC unit.

I don't understand it but at least I'm finding charts that show it.

[Sandy]

Looks like Power Factor Correction might be what is causing this issue.

The TIG 375 (newer) manual shows an idle current draw of 62 amps with PFC.

In the Miller 250DX manual I found a chart that shows that machine pulling 45 amps at idle with PFC but only 3 without PFC. At 60% duty cycle the PFC equiped machine pulled 61 amps to the 82 amps of the non PFC unit.

I don't understand it but at least I'm finding charts that show it.

Very interesting, very. I am as far from knowledgeable on power factor issues so things like this are a lot of food for thought.

As expected there is always a certain amount of loss and ineffeciencies between the power we receive and the power we use, thus the need for PFC. From looking at the current pulled during load conditions it shows what PFC can do for you, 61 amps vs 82 amps for the same amount of output.

Now according to all the explanations I've seen, which isn't a lot, the current pulled by PFC isn't a current that spins the meter and costs money. Sloshing back and forth are the words I see. I don't know how that can be but there is a lot I don't know or understand.

But when considering any negatives that might come from the addition of PFC capacitors, this is definitely a big consideration. It could take a welder out of the 'exception' catagory as far as the NEC is concerned and put it into the 'continuous duty' category like any other item.

I say this because welder circuits are allowed to be derated in size based on the assumption that the rated current draw is seldom long period let alone full period (continuous duty), ie only during the arc/welding mode. In between the draw is only minimal for fans, PCB draw etc.. If a welder with PFC is going to be pulling 45 to who knows amps, heating up wires and tripping breakers while merely 'turned on' all the wiring is going to need to be sized accordingly just like an electric heater or any other heavy draw piece of equipment.

Seems like a guy needs to think of all the +'s and -'s and decide whether it's really worth the hassle or not.

This is a lot of rambling so jump in and correct me where I'm off in my thinking.

Thanks for posting your results. Many don't and we are left to wonder what the heck ever happened.

[Nobody]

Seems like a guy needs to think of all the +'s and -'s and decide whether it's really worth the hassle or not.



If I had known this going into it I probably wouldn't have bought a unit this big, as it goes it was fairly cheap and local. I'm guessing that the primary design intention for this machine was for an industrial weld shop with a lot of power on tap and little idle time on the machines... pretty much the opposite of home shops. (At least my home shop)

I noticed in the manuals for the 25x series machines that the PFC capacitors are an optional kit that can be added on but come standard on 35x machines, this makes me wonder if I can get away with unhooking the caps that make up the PFC on my machine. Aside from possible unwanted harmonics in the transformer unhooking the PFC caps might cause some light flicker through out the house while welding... or so my electric arc furnace buddies tell me.

So anyone in Los Angeles wanna trade me for a smaller unit? :p

[Sandy]

I noticed in the manuals for the 25x series machines that the PFC capacitors are an optional kit that can be added on but come standard on 35x machines, this makes me wonder if I can get away with unhooking the caps that make up the PFC on my machine. Aside from possible unwanted harmonics in the transformer unhooking the PFC caps might cause some light flicker through out the house while welding... or so my electric arc furnace buddies tell me.

It seems like it's worth checking into anyway. Especially if it can be disabled without altering the wiring connectors and such. Maybe like laying both leads on one lug of the cap or something like that. That way it could be restored to the original state for sale or trade. Maybe the initial surge wouldn't too bad if you're mostly at the low end anyway. Hopefully most of the filtering for hamonics is on the far end of the circuitry.

If it was a real good deal you could get some more comfortable usage out of it and still be shopping around and come out in good shape.

[Nobody]

Now that I'm confident that this is a normal condition for this unit I'll be paying a visit to the electrical department for some heavier wire; I figure hardwire with 6ga to the panel on a 60A breaker should be sufficient for now along with turning it off when I run in the house or get talking with buddies.

The caps are easy to take offline as they are connected with spade lugs that can be disconnected and taped. Just for kicks I'll make another inquiry with Lincoln to see if disconnecting the caps is an acceptable modification for use in an low production / low current environment... no telling if they are there for more than just PFC so I would rather be safe than :cry:.

I appreciate your feedback; cool forum, I think I'll be spending some time around here.

[JohnnyWeld]

Just thought I'd chime in on that Lincoln 355. I have the same machine , and before I had my shop, I used it in my garage where the same thing happend(60amp breaker wasn't making it). Mine is rated 100amps 230v 1phase. If I were you , Nobody , I wouldn't think about selling that machine, especially if you got it at a good price. By the way, I'm pretty sure mine was manufactured late 90's .

[Nobody]

I used it in my garage where the same thing happend(60amp breaker wasn't making it). Mine is rated 100amps 230v 1phase. If I were you , Nobody , I wouldn't think about selling that machine, especially if you got it at a good price. By the way, I'm pretty sure mine was manufactured late 90's .

Interesting info on the 60amp breaker performance, if that won't cut it then it is going to be cost prohibitive for me to upgrade all of my electrical (again) just to support this unit. Was the breaker tripping at idle or while welding? Do you happen to know the type of circuit breaker in use?

I find it interesting that my 50amp breaker will survive for 5+ minutes at time, it's a bottom of the line square D HOMeline breaker that I'm lead to believe is a thermal trip as opposed magnetic trip breaker which may explain why it will take a bit of a temporary overload.

Kind of funny that when I was looking at TIG welders the same place had a used Lincoln 25x series machine... the only reason I passed on it was that it had all soft touch controls and I happen to like (and understand) knobs and dials as opposed to screens and menus.

[smithboy]

Is it possible to wire directly into the main breaker safely?

...I guess this question is directed at sandy.

[BBchevy396]

Hey guys, I read through the posts, and dont really have anything useful to add, so forgive me. But, I find it bewildering that such a large amount of energy is being consumed, with no work being done.
Kind of like your truck gets 15 MPG on the hwy, but uses 10 gals per hour while its off, sitting in your driveway!
I have a decent understanding of electricity, but somthing seems very odd, that this is "supposed" to be such a grossly inefficient machine.
if it is tripping breakers, and making heat in the wires, and an amp gauge reads load...... Its burning juice baby!!! and Dollars!!!! How could that be better? or right?

Just had to add..... does it have a "leak"?... perhaps a short to the case?

[Nobody]

I have a decent understanding of electricity, but somthing seems very odd, that this is "supposed" to be such a grossly inefficient machine.
if it is tripping breakers, and making heat in the wires, and an amp gauge reads load...... Its burning juice baby!!! and Dollars!!!! How could that be better? or right?

Just had to add..... does it have a "leak"?... perhaps a short to the case?

Actually it's not buring dollars... the power meter doesn't move noticably faster with the machine turned on and idle, if I start welding and using 'real' power then the meter will account for it.

From Lincoln...

Good afternoon. The high input current is the result of power factor capacitors accross the input. Actual input Watts, which is what you are charged for is much less, and at rated load, the input current will remain low.

http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html

Has some good info... makes a little more sense mathmatically than just sitting down thinking about it with an amp meter.

I hardwired to the panel with 6GA SO cord and thus far haven't had any problems... the breaker still gets warm but I shut the machine down between welding setups.

I'm going to keep my eyes peeled for a smaller machine and probably sell this one but for now I'm f**kin hooked this is my first experience with TIG welding and I love it!!! :cool2:

[Sandy]

Is it possible to wire directly into the main breaker safely?

...I guess this question is directed at sandy.

In general no it isn't.. Their design is such as to deter just that.

[awright]

I'll venture a comment here, despite the fact that I am a mechanical engineer, not an electrical engineer, and do not claim to be an expert on power engineering.

What you are measuring on the input line to the welder at idle with the clamp meter is REACTIVE current because the welder looks like a (almost) pure inductor across the line when it is idle. If you had a truly pure inductor across the power line, the current would be exactly 90 degrees out of phase with the power line voltage, you would be drawing no real power, and your electric meter would not be turning. In fact, there are always some losses in real, not ideal, equipment, so there is always some real power being drawn by the idle welder. Some of the loss is due to heating of the wiring, some goes into heating the magnetic core of the transformer due to eddy currents and magnetic excitation in the steel, and some is used to power the controls and accessories.

The power company does not like reactive loads on its lines because (among other problems they cause) the reactive current is real current that causes losses and heating in its transmission and distribution equipment while they are not getting paid for it because your meter is not turning. In fact, industrial users have to pay a surcharge if their demand is reactive, which is the primary motivation for installing power factor correction capacitors.

A pure CAPACITIVE load across the line also draws reactive current that is 90 degrees out of phase with the line voltage. If you had large, pure capacitors across the line you would also be drawing current without turning your meter.

The difference between INDUCTIVE reactive current and CAPACITIVE reactive current is that the former LAGS the line voltage, while the latter LEADS the line voltage. (You can remember which is which by thinking of the capacitor as a bucket that has to be filled with electrons. The current has to flow to fill the bucket BEFORE the voltage across the capacitor can build up, so the current has to LEAD the voltage.)

Now, in order to optimize the POWER FACTOR of the welder, the manufacturer places capacitors across the transformer primary so that the capacitive reactance will balance out the inductive reactance, tending to bring the power factor closer to unity, the ideal condition. Unity power factor occurs when the current drawn by the load is exactly in phase with the line voltage, all power drawn by the load is REAL power, not REACTIVE power, and (hopefully) real work is being done by the power. This balancing can only be accomplished perfectly at one operating condition, but under most operating conditions, the deviation from ideal power factor will be less than it would be in the absence of the power factor correction capacitors, making the power company happier (and you don't want to make the power company mad, because they will get even).

Back to the original question, Nobody asked, "... why is this thing pulling 65 amps no load? Why does it pull less power from the grid as the power to the work increases?" Well, as some of the responders have indicated, the welder is NOT pulling less power from the grid as load increases. To measure true POWER, you must measure the part of the current that is IN-PHASE with the voltage and multiply that current by the voltage. You can't do that with a plain clamp-on ammeter and a voltmeter. You can do it with a WATTMETER that accomplishes just that in-phase multiplication internally (originally by having a meter movement with a heavy coil to energize the meter magnet with the load current instead of a permanent magnet, while the voltage was impressed on the moving coil of the meter, but more recently by electronic manipulation). The electric company does it with an electric meter that also does the multiplication of in-phase voltage and current.

BBchevy396 says, "But, I find it bewildering that such a large amount of energy is being consumed, with no work being done". Well, a large amount of power is NOT being consumed with the machine at idle, as a look at your power meter will confirm, because most of the current you measured with your clamp meter is reactive.

Hope I have not added to the confusion.

awright

[TxRedneck]

awright, bravo, for those who arent thoroughly confused I believe they understand quite well. My granddad is an EE and although wont bother coming onto a computer to look at stuff like this....somehitng about his eyes he says, reading this was like listening to him. Does make sense. I did enjoy reading this and do appreciate you taking the time to expalin this. Also, ID like to welcome you to the board :D Im sure we could use a good engineer on more than a handful of projects and ideas folks have.....Welcome:waving:

[12,000 Doors]

Is it possible to wire directly into the main breaker safely?

...I guess this question is directed at sandy.

It may not be safe but I've done it a whole bunch of times. When installing gates and railings at peoples homes that were fed 220v but didn't have any recepticals I would loosen up the three set screws that tightened on the three wires coming off the pole at the top of the breaker box. I would work one at a time and after I loosened each one I would jam one of the three wires from my welder in that hole along with the wire from the pole and tighten it back up being carefull not to touch any other wires. Ofcoarse I made sure my green wire went with the green wire from the pole (the common ground). In hindsight it probably would have been a lot safer if I had a breaker on the cord to my welder. Wear good gloves just in case. Your working without a net (breaker) if you take this on. I've always been real carefull and a little lucky.

[Nobody]

Thanks for the replies y'all.

I've found PFC to be somewhat of a mystery even within the power engineering communities; folks either understand it compeletly or don't have a clue about it.

Thus far I'm surviving on a 60amp breaker with #6 SO cord, the cord doesn't get hot anymore but the breaker still heats up, I usually shut things down when it gets up into the high 90s or I won't be welding for a few minutes.

Normally I would come off a 100amp breaker but the wiring between the main and sub (#6) wouldn't handle it... I'm not sure about 60amps but thus far the wiring hasn't gotten hot at the parts I can measure.

I'm still in the market for a smaller unit... just haven't found anything that looks like a good fit for me.

[TxRedneck]

I was on the aws forum and read a similiar question. Similar is the operative word. The youngster had a tig welder machine and a generator of I dont remmber what size. He found the generator engine to labor dramatically. He was told do add a PFC by way of capacitors. They told him any good electrician could do this. It was deemed that while he was withen the KW's he wasnt withen the KVA's...or atleast that what I think I had read. But the 64 million dollar question was, if I understand this correct enough to ask, I beleive y'all said the power thats reactive would go back tothe power company?? well what about on a generator? Is this just lost energy? I understand from physics you cant just destroy engery, or whatever. That in all cases when energy changes from one form to another there is some waste, which becomes loss due to waste, often in the form of heat...which is all the while still energy. So have I basically answereed my own question? are we seeing the energy just turned into work loss due to heat?

[Sandy]

He was told do add a PFC by way of capacitors. They told him any good electrician could do this. It was deemed that while he was withen the KW's he wasnt withen the KVA's...or atleast that what I think I had read.

Without reading that post, I'm going to say what you were seeing there was discussion about the need for and the positive apsects of power factor capacitors (PFCs). As always there are one or more reactions to any given action, some good and some bad depending on the desired outcome.

Due to a lot of things, we'll call them losses and inefficiencies, the difference between the power sent to a tool, motor, or any utility type appliance such as a welder and the power that actually gets used by that item to produce real work can be significant. Sometimes as much as 20%. There is a number used to quantify that. It is called the Power Factor or Pf. The Power Factor for a motor might be, for example, point 84 (.84). Which means that about 84% of the power that this motor called for got used in and through the motor and 16% got lost. Power Factor Capacitors can help narrow this gap.

The case you describe is one where a PFC might be usefull. When something as a gen set is marginal in it's abilities to supply the power required to supply this welder, PFCs just might be able to squeek out that little bit extra by narrowing the Power Factor to maybe, ummmmmmmm point 95. There are charts for correctly sizing PFCs.

With a couple of negatives in mind I dunno if that would be my long term answer for an undersized gen set.

[Sandy]

I need to add that the numbers I threw out should by no means be taken for any other case than demonstrating the Power Factor scenario.

[TxRedneck]

ill try to find that thread and post the post here

[TxRedneck]

Topic mobile welding off a genorator?


Go to previous topic Go to next topic Go to higher level

By acs welding On 26-Oct-05 16:22


[Reply]

i got abot 6 welders they all work real well off my genorator which is seperate unit however i just bought a miller syncrowave 180 sd
and my buddy told me i will only be able to stick weld with it up to about 130 amps and i can forget about tig welding off generator unless i use lower amps and scratch start! is this true i wont be able to use foot pedal when mobile tigging? i mig and stick off it ok the generator is 10,000 watt
13,500 surge watts also is there a batery pack i can buy to give me constant current?


By BillC On 26-Oct-05 17:38


[Reply]

I did some steel GTAW last weekend with my 180SD plugged into my 10kW Trailblazer. I used the foot pedal and had the current set at 120 amps DC. Worked fine for me...

Regards,


By metallord On 26-Oct-05 18:01


[Reply]

the high freq is built in on a 180 sd.if the machine is running the h.q. is working.


By acs welding On 26-Oct-05 18:41


[Reply]

i said im fine up to 130 amps ..tried it at 180 amps and generator
sounds really loaded almost bogged down to where it wanted to stall
maybe it works better with trail blazer ..i was told i cant use the 180sd at all with a generator because its solid state ? i tried it with stick worked great with the stick up tp 130 amps...havent tried tig yet dealer told me ill be fine as long as input voltage doesnt drop below 208 volt... im confused here


By OSUtigger On 26-Oct-05 23:11


[Reply]

A 10kW generator *should* produce the same output as a wall outlet, 230 Volt, about a 40 amp breaker. I cannot see why this is not the case, but if someone does, please correct me. Yes, the engine will bog. It is sized to put out that amount of power. You engine in your vehicle does the same thing when you accelerate. As long as it's running, it can handle it, no matter the sound the engine is making. Generators are also designed to put out a constant voltage, and I will bet that if that voltage drops too much it will simply kick a breaker or reset button on the generator itself. You do not need a constant current source (if you had a constant current source, you would not need the stick welding option on the 180SD, only the TIG high freq option), nor any accessories to create a constant current source. The only reason that you could not use the TIG to its full extent would be if for some reason the welder is pulling quite a few amps to do something else other than just the welding (for example, running a pump for the water cooled torch if you have one, or perhaps a grinder or chop saw by someone else on the crew).

If this is not the case, either something is wrong with your generator or I have a completely misconstrued view of how generators actually work, and if so, again, I hope to be corrected.

Hope that helps!

gls


By billvanderhoof On 27-Oct-05 01:51


[Reply]

Going a bit out on the limb here, it's been a long time since that AC dircuits course. I looked at millers specs and they say at 150 amps it draws 6kw but 12.3kva. That indicates a quite reactive load and I believe your generator sees the 12.3kva as the load. The difference has to do with the fact that the voltage and amps are not in phase with each other when the load is reactive. It seems that your finding that the generator can only power it up to about 130 amps may be correct. I don't see why TIG would be different. That is works ok at lower settings. Experiment.
bill


By acs welding On 27-Oct-05 05:36


[Reply]

i started my bussiuness about 4 months ago while i was still in welding school getting my aws certs when i was
shoping for genorators o had cal;led lincon electric
they told me volts x amps=watts now i added all that up and i would need 6,900watts to run it ar 160 amps in stick mode however
the lincoln invertec v 160s also only runs about 140 amp max
i was under the impression my self that it was same as wall outlet
but a couple of rhe instructors are telling me its not the same
as the wall outlet has constant power the generator doesnt so when i ask it for more through the foot pedal it won be there
sure hope they are wrong
thanks for help andy


By OSUtigger On 27-Oct-05 13:49


[Reply]

Andy,

Bill pinned it right on the nose. For some reason, the TIG has some sort of inductance in its circuit, and therefore it runs like a motor. You will find that if you go to an electrician and have him install a capacitor somewhere in the circuit (he should know what this means), the machine *can* work perfectly, or, depending on the skill and judgement of the electrician, about 90% efficient (this is called power factor correction, and has everything to do with the difference between kVA and watts). Sorry this didn't hit me before. You are actually pulling about 12.3 kW from the generator, but the machine is actually only using 6.9kW, therefore you have a PF of about 50%. An electric company will ask that you have the same thing done when you install a large single phase motor, because they can only charge for real power, and not for kVA.

Hope that helps.

gls


By acs welding On 28-Oct-05 08:48


[Reply]

sweet..ok that was a bigg help!! as i have a real powerfull car stereo my amplifiers in the truck have capacitors which hold juice so head lights dont dimm when bass hits and i needed this even when i upgraded
stock altenator from 135amp to 220 amp any way where i was goining with this was is there a battery pack or capacitor that i could use to give me more steady power? and you just answered that for me
great guys thanks for all the help so basicly the capacitor or batt pack would go between the gen and welder?and i would get juice from batt pack or cap and gen gharges those up correct?


By BillC On 28-Oct-05 11:01


[Reply]

No, the capacitor is not acting as a big battery pack. It is changing the phase of the AC. Do a google search on "power factor correction"...


By OSUtigger On 28-Oct-05 11:51


[Reply]

Again, Bill's right, you absolutely have to get the exactly right size of capacitor, and though it is storing electrons, it IS NOT a battery. It is simply realigning the sine waves. This works completely differently than how your capacitor is working for your stereo. The capacitor for your stereo is storing the electrons for an instantaneous moment when they are needed, and if you keep bumping the speakers continuously, eventually you will not be able to store the capacitance necessary to fully bump them. A capacitor in an AC cicuit more or less delays or advances voltage according to time, and makes it so that you don't have a case of peak amperage at minimum voltage or vice versa. It is hard to explain, and I don't remember everything about it like I should, but it involves real as well as "imaginary" numbers (square root of -1 stuff).

Point is, you can't just throw a big ol' capacitor in and expect it to work. You have to get the right size, and this requires quite a bit of evaluation.

gls


By acs welding On 28-Oct-05 22:15


[Reply]

ok i get it im gonna have a tech do this and have it calibrated too
you and billc were a big help was getting ready to buy a bigger genorator


By billvanderhoof On 29-Oct-05 01:34


[Reply]

As a side benefit the power factor correction will save you fuel.
Bill


By BillC On 29-Oct-05 08:39


[Reply]

Bill,

That brings up an interesting question... The inefficiency is current draw, but not power consumption. When you are hooked up to the local power company, you are continuously drawing more current than you need during half a cycle, then returning the excess to the grid during the second half of the cycle. What happens to that excess current with a generator?

hope this makes more sense

[Sandy]

TxRedneck,

I liked that last question there. Let me know the unofficial answer. :)


I would tend (opinion only) to disagree with the comment about saving fuel. This is one reason I try to stay away from discussions on power factor capacitors==one billion variables. Again, the reason for installing PFC must be to acheive very specific results for one reason or another or one party or another.

I think that in the case of a gas powered generator, the PFC is going to place an apparent (dummy) load on the generator, especially during the tig idle periods, and cause the motor to actually run at more of a load than it would under the normal "short weld"--"long pause" conditions that we usually create. That's should consume more gas as under normal conditions the engine usually gets a little breather between weld cycles as in weld--pause weld--pause and so-on. With PFC it won't get any breather. The largest savings would be if one were able to keep the tig working in the moderate to upper ranges of its capacity for most of the gen run time. That's where the power thus load (gas) savings would be. That ain't likely.

Like I said before, we put in PFCs for specific reasons at different times. In this case it would be to squeeze every last electron out of an otherwise undersized gen set. I wouldn't hope for fuel savings at the same time.

Here's a little more food for thought. PFCs do a great job in the right place at the right time BUT if Power Factor Capacitors were the answer to all our woes, they would be standard wiring on every piece of 230 equipment we buy. They ain't. :)

We'll wait and see just how far off I can be here.............. :laugh:

Sorry for the excess edits. Spelling and rambling mind.

[TxRedneck]

Sandy, good points, specially on gas consump. I will try to keep up on this thread hoping to get the answer :D

[caosesvida]

good posts here, very interesting.
BUT 12,000 doors, that is really a very bad practice to be doing with the main lugs and loosening them, Under certain circumstances you could get a very serious arc fault going there, which would give you 3rd degree burns all over your face and arms, if you weren't killed by the fire ball going down you lungs. I know it sounds dramatic but it is a fact of life. This can happen even if you are not fooling around with the wires, I have seen panels blown off the walls and a complete meltdown of a hotel mechanical room because of this. Please be safe and try to find another option, such as taking out a breaker temporarily and putting in your machine on that breaker, its not that much more time consuming as you have the cover off, they don't need the stove or hot water heater for the time you are welding, and its much more safe, in fact may be legal. good luck.

[gocirino]

Found this regarding power factoring;




AMPS, WATTS, POWER FACTOR AND EFFICIENCY
WHAT DO YOU REALLY PAY FOR?
INTRODUCTION

There seems to be a great deal of confusion among the users of electric motors regarding the relative importance of power factor, efficiency and amperage, as related to operating cost. The following information should help to put these terms into proper perspective.

At the risk of treating these items in reverse order, it might be helpful to understand that in an electric bill, commercial, industrial or residential, the basic unit of measurement is the kilowatt hour. This is a measure of the amount of energy that is delivered. In many respects, the kilowatt hour could be compared to a ton of coal, a cubic foot of natural gas, or a gallon of gasoline, in that it is a basic energy unit. The kilowatt hour is not directly related to amperes, and at no place on an electric bill will you find any reference to the amperes that have been utilized. It is vitally important to note this distinction. You are billed for kilowatt hours: you do not necessarily pay for amperes.

POWER FACTOR

Perhaps the greatest confusion arises due to the fact that early in our science educations, we were told that the formula for watts was amps times volts. This formula, watts = amps x volts, is perfectly true for direct current circuits. It also works on some AC loads such as incandescent light bulbs, quartz heaters, electric range heating elements, and other equipment of this general nature. However, when the loads involve a characteristic called inductance, the formula has to be altered to include a new term called power factor. Thus, the new formula for single phase loads becomes, watts are equal to amps x volts x power factor. The new term, power factor, is always involved in applications where AC power is used and inductive magnetic elements exist in the circuit. Inductive elements are magnetic devices such as solenoid coils, motor windings, transformer windings, fluorescent lamp ballasts, and similar equipment that have magnetic components as part of their design.

Looking at the electrical flow into this type of device, we would find that there are, in essence, two components. One portion is absorbed and utilized to do useful work. This portion is called the real power. The second portion is literally borrowed from the power company and used to magnetize the magnetic portion of the circuit. Due to the reversing nature of AC power, this borrowed power is subsequently returned to the power system when the AC cycle reverses. This borrowing and returning occurs on a continuous basis. Power factor then becomes a measurement of the amount of real power that is used, divided by the total amount of power, both borrowed and used. Values for power factor will range from zero to 1.0. If all the power is borrowed and returned with none being used, the power factor would be zero. If on the other hand, all of the power drawn from the power line is utilized and none is returned, the power factor becomes 1.0. In the case of electric heating elements, incandescent light bulbs, etc., the power factor is 1.0. In the case of electric motors, the power factor is variable and changes with the amount of load that is applied to the motor. Thus, a motor running on a work bench, with no load applied to the shaft, will have a low power factor (perhaps .1 or 10%), and a motor running at full load, connected to a pump or a fan might have a relatively high power factor (perhaps .88 or 88%). Between the no load point and the full load point, the power factor increases steadily with the horsepower loading that is applied to the motor. These trends can be seen on the typical motor performance data plots which are shown in figure 1.

EFFICIENCY

Now, let’s consider one of the most critical elements involved in motor operating cost. This is efficiency. Efficiency is the measure of how well the electric motor converts the power that is purchased into useful work. For example, an electric heater such as the element in an electric stove, converts 100% of the power delivered into heat. In other devices such as motors, not all of the purchased energy is converted into usable energy. A certain portion is lost and is not recoverable because it is expended in the losses associated with operating the device. In an electric motor, these typical losses are the copper losses, the iron losses, and the so-called friction and windage losses associated with spinning the rotor and the bearings and moving the cooling air through the motor.

In an energy efficient motor, the losses are reduced by using designs that employ better grades of material, more material and better designs, to minimize the various items that contribute to the losses in the motor.

For example, on a 10 HP motor, a Super E energy efficient design might have a full load efficiency of 91.7%, meaning that, at full load (10 HP), it converts 91.7% of the energy it receives into useful work. A less efficient motor might have an efficiency of 82%, which would indicate that it only converts 82% of the power into useful work.

In general, the efficiency of motors will be relatively constant from 50% to 100% of rated load.
AMPERES

Now, let’s discuss amperes. Amperes are an indication of the flow of electric current into the motor. This flow includes both the borrowed as well as the used power. At low load levels, the borrowed power is a high percentage of the total power. As the load increases on the motor, the borrowed power becomes less and less of a factor and the used power becomes greater. Thus, there is an increase in the power factor as the load on the motor increases. As the load continues to increase beyond 50% of the rating of the motor, the amperage starts to increase in a nearly straight line relationship. This can be seen in figure 1.

SUMMARY

Figure 1 shows significant items that have been discussed as plots of efficiency, power factor and watts, as they relate to horsepower. The most significant factor of all these is the watts requirement of the motor for the various load levels because it is the watts that will determine the operating cost of the motor, not the amperage.

The customer that has an extremely low power factor in the total plant electrical system, may be penalized by his utility company because he is effectively borrowing a great deal of power without paying for it. When this type of charge is levied on the customer, it is generally called a power factor penalty. In general, power factor penalties are levied only on large industrial customers and rarely on smaller customers regardless of their power factor. In addition, there are a great many types of power customers such as commercial establishments, hospitals, and some industrial plants that inherently run at very high power factors. Thus, the power factor of individual small motors that are added to the system, will not have any significant effect on the total plant power factor.

It is for this reasons that the blanket statement can be made, that increasing motor efficiency will reduce the kilowatt hour consumption and the power cost for all classes of power users, regardless of their particular rate structure or power factor situation. This same type of statement cannot be made relative to power factor.

The following basic equations are useful in understanding and calculating the factors that determine the operating costs of motors and other types of electrical equipment.

OPERATING COST CALCULATIONS

MOTORS

Kilowatt Hours = (HP** x .746 x Hours of Operation)/Motor Efficiency

** Average Load HP (May be lower than Motor Nameplate HP)

General Formula All Loads

Kilowatt Hours = (Watts x Hours of Operation)/1000

Approximate Operating Cost* = Kilowatt hours x Average Cost per Kilowatt Hour

* Does not include power factor penalty or demand charges which may be applicable in some areas.

Useful Constants

Average Hours per Month = 730

Average Hours per Year = 8760

Average Hours of Darkness per Year = 4000

Approximate Average Hours per Month(Single Shift Operation) = 200



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[TxRedneck]

great :) thanks gocirino

[Sandy]

gocirino,

You're hurting my head posting all that at once.. :)

I can handle about one detail a day.

[awright]

Thanks for your kind remarks, TxRedneck. Yeah, gocirino, that's an impressive contribution. I haven't read it yet, but printed it out to digest later.

A subtext of this thread is, "what's in it for me to correct power factor on my machine?" Well, part of the answer has to do with what could be called good citizenship. (That's not meant to be snide. I can't think of a more apt term at 2 am.)

By pure coincidence, the January '06 issue of the IEEE (Institute of Electrical and Electronic Engineers) magazine, "IEEE Spectrum", that arrived yesterday contained a long article on a very expensive power factor correction device using superconducting technology and exotic control electronics that is used at the utility level. I haven't read the whole article yet, but skimming it, one paragraph caught my attention:

"Reactive power (provided by the new device - awright) is what's required to counteract lagging currents and sagging voltage (on the grid due to customer's inductive loads - awright), and if it isn't supplied quickly or efficiently enough, networks crash or equipment suffers. Indeed, reactive-power-supply problems are among the chief culprits in an overall power-anomoly and -disturbance problem that costs the United States alone between US$119 billion and $188 billion a year in lost economic activity, according to a 2001 report by the Electric Power Research Institute, in Palo Alto, Calif. Such staggering losses add up to 1.2 percent to 1.9 percent a year of the country's gross domestic product, the report noted."

You can see why the electric utilities penalize industrial customers that put large reactive loads on the grid.

So, it's kind of like voting - your individual action is small, but is an important part of the big picture. The day may come soon when the use of power factor correction is mandatory on welders, as is now the case with some consumer electronics power supplies (which are individually an infinitesimal part of the overall problem, but cumulatively are becoming a major impact on the grid).

I don't know about saving gas, but if you are using an auxiliary generator to supply your line-operated welder, power factor correction may save your generator from damage. It doesn't like reactive loads any more than the utility does.

Note that the welding arc itself is not reactive - it is essentially resistive, once struck, which means non-reactive. It is the non-ideally loaded transformer and current control reactor (coil) that causes the reactive load. (Now, here's where I stick my neck out, since I've never used or looked inside an engine driven welder.) I'm guessing that purpose-built engine-driven welders do not contain large transformers, since voltage/current control are achieved by generator exciter control, rather than transformer tapping or adjustment, and do not, therefore, suffer from severe reactive load problems. Is that true?

Incidentally, I totally concur with caosesvida's cautionary comments to 12,000 Doors. The grid can supply an enormous surge of power that can kill or maim instantly and only the slightest lapse of attention or lost grip on a cable can be disasterous. Tapping power downstream of the main panel supply breaker or switch is MUCH less hazardous and allows connections to be made with the panel disconnected from the grid. By law, there has to be an accessible disconnect at the panel.

awright

[TxRedneck]

AWright, I cant comment on the particular electrical aspect of the welding generator cause im an electrical idiot. I understand that there are in hence two types of weldin ggenerators. One is the traditional pure dc generator. This isnt what were refering to. The other, the ac generator, or I think is an alternator actually, and rectified to dc. Well the point being is that Ive been told that these machines, and the stand alone generators are actually not quite the same. That the welder generator does suffer some performace loss due to differnces in the characteritics which are needed for making a welding arc verus giving electrical power. I guess there maybe something to do with the low volt high amp converison? I dont know. anyway, next time I see him ill ask GD to take a look at my bobus and tell me what he says. Dont know when thatll be though . Thanks for all the good info guys

[broberts5]

I have the same Lincoln TIG 355 and wouldn't trade it for the world. What a great welder! I am only running a 100 amp breaker on it and have had no problems. I do a lot of aluminum tig at 230 or better and, on ocassion, have been around 300 amps with no breaker meltdowns. I weld for a living and don't really see much difference in my bill whether I am running the welder a lot or hardly use it al all.

[castirondude]

hoowah I just bought a lincoln 355 that came from a company, it has power factor correction and sure enough it does the same thing - 70 A load at idle! I have it on a 100A circuit so it does OK but I can actually hear the buzzing in the house. It seems better to have half the capacitor bank to get a middle ground.

[castirondude]

By the way, I am an electrical engineer and studied AC electronics and power factor correction in college.

For those of you who don't have an electrical / math degree, a good analogy would be balancing a tire. In the case of a balanced tire, the wheel bearings and springs only carry the load of the vehicle. When your tire is out of balance, the tire bounces up and down - when the tire goes down the load on the bearings is temporarily reduced, and when the tire moves up, the load is temporarily increased. On average, the load is the same, but the peaks are much higher. Even if your truck was sitting on jack stands with the tire in the air, the tire would vibrate and cause an artificial load on the bearings (be it alternating positive and negative, average zero)

The latter is exactly what is happening electrically with the 70A idle current. You're stressing your electrical system by drawing power (stored in the capacitor bank) and delivering it back 60 times a second, averaging close to zero wattage (load).

A resistor is a perfectly balanced load. An inductor or capacitor is an unbalanced load and in the case of taking 70A without burning power, realistically it is alternating a positive and negative load, averaging close to zero. With the electric company you only pay for the actual load you put on the net, the rest is overhead mostly on their tab and they don't like that. I think you'll find you are paying some of it as well though.