Power factor caps and how to calculate them?

Gunner · 2004-12-14T06:12:36+00:00

Any idea of how to calculate the amount of capacitance needed toreduce 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

D

dcaster 21 years ago

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,

Vote

D

dcaster 21 years ago

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.

measurement

current

Vote

G

Gunner 21 years ago

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

Vote

J

JK 21 years ago

Actually, this is starting to piss me off. Are you saying that I can put 50 - 70 uF across L1 and L2 in a grey box and it is the same as what Miller wants to sell me for $200 usd?!

Regards, Jim

Vote

D

dcaster 21 years ago

I have wondered about the high prices charged for adding PF correction, and keep thinking there must be reason for the high price. That is reasons beyond that Miller would be buying new caps and reselling them at a profit, and then the distributor would be wanting to make a profit too. There might be some reason beyond that they can and do sell some for $200 usd. , but I sure don't know what it is.

Actually I would never recommend putting 50 - 70 uf in a grey box. I am too cheap for that. I would see if there wasn't some space in that big red box you already have. That is what I did with my Buzz box years ago.

Dan

JK wrote:

Vote

D

dcaster 21 years ago

I did say to measure the current with no PF correction and with some PF cap installed and I would help you with the rest. Let me now amend that to those two measurements and also the current to just the PF cap. I will help you understand it all if you get those three measurements.

Dan

Vote

J

JK 21 years ago

Dan,

I routinely use a Miller 251 wire feed or a Miller 250 DX Tig machine.

Is there a ball park figure to use a for the power factor correction capacitors that I missed somewhere. I built my own 7.5 to 17.5 hp phase converter to run my shop equipment so I understand the dangers associated with 220 v. I tuned it to optimize performance with balancing capacitors.

regards, Jim

Vote

D

dcaster 21 years ago

<

formatting link

> is the first site I found by searching on Google for " Power Factor correction ". It give you some rules of thumb for PF correction for motors and some idea of why one does not try to get 100 % power factor. Take a look there and come back and ask if you have questions.

Dan

Vote