Power factor caps and how to calculate them?

Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder?
Gunner, Idealarc Tig250/250 and Miller Dialarc 300

"To be civilized is to restrain the ability to commit mayhem. To be incapable of committing mayhem is not the mark of the civilized, merely the domesticated." - Trefor Thomas
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Gunner wrote:

Calculating isn't easy. Much easier to add some capacitance and check current, add some more, keep testing until you find a minimum idle current.
I believe that for optimal performance you would want SQRT(2*PI/LC) = 60 Hz.
You know everything in that equation except the C. That assumes you know the effective inductance "looking into" the primary of the transformer.
I think that an LC network looks resistive at the resonant frequency, and that the resonant frequency is SQRT(2*PI/LC). That's why if you add just the right amount of capacitance for the inductance of your welder, then the resonant frequency is the same as that of the power grid and you get zero imaginary current; i.e. all of your current will be real current.
Real current and imaginary current are electrical engineering terms. Imaginary current can burn up a wire just like real current, it just doesn't show up on your power meter.
I used to know all this stuff ..
Grant
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Hz.
Er, F = 1 / (2 * pi * sqrt(L*C))

I suppose you could measure unloaded current draw. But some of that is actual power too...

Yep, the C and L cancel. But if you're blowing breakers, you've probably already got trouble simply turning it on. That power-on spike is none too fun for the circuit with *just* the inductance, let alone the caps (which look like a short to such transients)!

And I'm still learning it. :o
Tim
-- "I've got more trophies than Wayne Gretsky and the Pope combined!" - Homer Simpson Website @ http://webpages.charter.net/dawill/tmoranwms
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Would it be better to switch in the caps AFTER turn-on?
--
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On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin

Add it where, and how much do I start with? My power bill from Pacific Greed and Extortion (PG&E) was a $100 higher this month from what I suspect is the welders being used a fair amount..most of which was simply idling.

I never knew any of that stuff...shrug.

Gunnner
"To be civilized is to restrain the ability to commit mayhem. To be incapable of committing mayhem is not the mark of the civilized, merely the domesticated." - Trefor Thomas
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Gunner wrote:

Add capacitance right across the power lines, from L1 to L2; i.e. connect each capacitor in parallel with the whole welder.
I'd scrounge some 10uF motor run caps and add one and turn on your welder and measure the current draw (with a loop-type current meter) and then add another cap and measure it again. If it goes down then just keep adding capacitance. You might want to also add some 30 or 35 uF run caps just so you can build up quicker.
If you need help scrounging motor run caps drop me an email. I have an unimpeachable source.
Grant
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Grant Erwin wrote:

You want to measure the current on the meter side of the caps; not on the welder side.
--
Keith Bowers - Thomasville, NC

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Do you have one of the new "speed read" meters, or the old manual type? When there is a sudden jump in the bill it's always a good idea to check that the meter reading actually matches what the bill says.
Twice before they upgraded to the speed read meters I had to call in corrected readings when the meter reader misread one of the upper digits causing a 1,000 kwh jump in the bill. I probably didn't notice other errors in lower digits and fortunately the next correct reading will put things back on track anyway.
$100 jump would seem to be an awful lot of KWH, somewhere between 400 and 800 kwh depending on how bad your electric rates are. Even at 400 kwh that works out to on the order of 13 kwh / day more consumption *every* day during the month.
Pete C.
Gunner wrote:

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<<<snip>>>
Imaginary
on
If so, then balancing isnt going to lower the operating costs anyways.....
--

SVL




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This is true. Power delivered is power sold. IMO, it is highly unlikely Gunner used $100 worth of juice in one month just heating up the wiring during welder idle time. The extra charge is probably due to an error in meter reading - or - Gunner just did a lot more welding in that month.
Bob Swinney

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anyways.....
Yeah, I know......I just that I wanted third party verification for Gunner's sake.....
( Pretty sure he has me branded as being 'Liberal'--and so anything *I* might say is *definately* suspect )
<G>
--
SVL





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More like: Resonant Frequency = [1 / (2*PI*SQRT(L*C) ] and solving for C SQRT[1 / (*60*60*4*PI*PI*L)] But as stated you'd have to know the value of L . . . . .
Bob Swinney

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Gunner;     Have you checked to see if Miller sold Power factor Caps for your machine? Lincoln supplied the ones in my Squarewave 275. It was about $100 cdn at the time. Miller might give specs for theirs (if they sell them)
Pete
Gunner wrote:

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Pete Snell
Royal Military College
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Yes, I know how to calculate the required capacitor value, if you can give me the idling inductance of the transformer primary. But you don't know that, and neither to I.
I do know that Miller uses 250 uF of capacitance in their power factor correction kit for the Synchrowave 250. That makes it tempting to say 1 uF per amp of output, but that would be wrong. It depends on the particular transformer. Something in the neighborhood of 250 uF for your machines should be about right though. You don't have to hit it right on the nose to get a useful amount of PF correction.
Gary
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Hey Gunner,
For what it's worth, I brought home some oil-filled caps from Apex Surplus out in the Sun Valley area of LA. For my comments here, I'm making San Fernando as east-west, and Lankershim as north-south. Anyway, just in case you are NOT familiar with the place, it's on San Fernando just east of Lankershim. Finding anything in there on your own is a bit of a bitch though, but a real adventure. Ask at the counter if you're short on time. The larger ones I got were at the west end of the first or second row about waist high, and the smaller ones were on the north wall due north of the large ones (but over two rows). I haven't even unpacked mine yet though, so I don't know that they are "good" yet. The guy said I could bring them back if they didn't work, and he knew right away they were for a Rotary Phase Convertor.
Take care.
Brian Lawson, Bothwell, Ontario. XXXXXXXXXXXXXXXXXXXXXX
wrote:

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wrote:

Thanks Brian. Im well familiar with Apex. Someday I aspire to be 1% of them <G>
Sorry we didnt meet up this trip. I got your voice mail today. I was up on the roof in the rain..sigh.
Next trip, try to make a little time to come visit the homestead. Im sure we can find Stuff to fit in your suitcase <G>
Gunner

"The French are a smallish, monkey-looking bunch and not dressed any better, on average, than the citizens of Baltimore. True, you can sit outside in Paris and drink little cups of coffee, but why this is more stylish than sitting inside and drinking large glasses of whiskey I don't know." -- P.J O'Rourke (1989) --------------------------------------------------------------------------------
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Gunner wrote:

<snip>
Speaking of roof - house floating, dab of tar here and there or 5000 sq ft. tarp ? Hope it was simple!!!
Several years ago it was tar for me - I had a Coastal Redwood limb fall but end first and drive a hole through my shop roof. I found it the next day checking the house for leaks due to the massive wind storm and rain...
This time, Gunner gets the Pineapple express (sorry Hawaii! ) and I get a 24" oak to fall across my driveway across the 220 feeder lines and wedge into some Redwoods.
Thank the storm god on that one, it was the tree next to the power pole and the heavy weight - on the wires "threatened the pole" so the Power company PG&E called a tree service - Davey - managed to cut the 40 or 50' tree and drop or swing the chunks to the sloping driveway so they wouldn't roll. The driveway is between 30 degree and 45 degree slope depending on the side or location...
The shop is at the bottom of the slope (live on a hill side - 30' setback means down hill house...) and a bedroom and office in danger if the bottom end swings or kicks out and the top goes like a tossed log...
I got the tree down and then paid for a hauler to take just over 100 cu ft of cut and crushed down Oak and Redwood limbs to the dump. The logs sit for now - looking for a home.
I have a home for a few of them - anvil holders :-)
Martin (yet another near hit!)
--
Martin Eastburn, Barbara Eastburn
@ home at Lion's Lair with our computer snipped-for-privacy@pacbell.net
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Calculating the amount of capacitance needed to reduce the idle current is actually fairly simple. I have posted how to do it at least twice. Try seaching on google with "power factor correction snipped-for-privacy@krl.org "
What you do need to remember is that you really don't want to correct to more than about 85% power factor. Correcting for that last 15% will not change the idle current very much.
Okay, I will post this one more time. First measure the current drawn with the welder idling and no power factor cap. Now add a power factor cap right across the line in parallel with the welder. Measure the current again. Now for the graphic solution. Use a compass and swing an arc that is proportional to the first measurement from point A. Shorten up the compass and swing a second arc ( also from point A ) proportional to the current with the P.F. cap installed. Now subtract the second current from the first current and set the compass to this length. Now put the point of the compass on the first arc and swing an arc so it intersects the second arc. Draw lines from point A to where you had the compass on the first arc and from point A to where the intersection is with the second arc. DRaw a third line between the ends of the two lines. Now turn your drawing around until that last line is vertical. Draw a line from point A horizontal to underneath the vertical line. Extend the vertical line to the horizontal line.
Okay? The horizontal line from point A to the vertical line is the amount of current that is real. The vertical line is the imaginary current. The length between the two arcs is how much imaginary current flows through the P.F. cap. The line from point A anywhere on the vertical line is the total current. Note that doubling the imaginary current ( doubling the capacitance ) does not double the reduction is the total current.
This is not easy to explain. So measure the current with no P.F. cap and with a P.F. cap installed and post the results and I will do the graphics and tell you what I get for results.
Dan
Gunner wrote:

civilized,
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Why do I bother.......... No one has made any comments on what I posted.......... No even that it is Wrong, Wrong, wrong........
I did make a serious mistake. Where I said " subtract the second current from the first current " I meant to connect just the power factor cap across the power line and measure the current it draws.
But apparently no one read my post critically or else they would have said " WTF ". I am hurt........8-(.
Dan
snipped-for-privacy@krl.org wrote:

current
twice.
"
will
drawn
measurement
and
a
current
is
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On 15 Jan 2005 11:40:15 -0800, snipped-for-privacy@krl.org wrote:

Dan, I read it, but its over my head so I simply nodded, and printed it out to try in the near future.
Gunner

"At the core of liberalism is the spoiled child - miserable, as all spoiled children are, unsatisfied, demanding, ill-disciplined, despotic and useless. Liberalism is a philosphy of sniveling brats." -- P.J. O'Rourke
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