Home |
Search |
Today's Posts |
|
Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
Reply |
|
LinkBack | Thread Tools | Display Modes |
#1
|
|||
|
|||
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 |
#2
|
|||
|
|||
Gunner wrote:
Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? 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 |
#3
|
|||
|
|||
On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin
wrote: Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? 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. 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 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 .. I never knew any of that stuff...shrug. Grant 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 |
#4
|
|||
|
|||
In article ,
"Tim Williams" wrote: 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)! Would it be better to switch in the caps AFTER turn-on? -- Free men own guns, slaves don't www.geocities.com/CapitolHill/5357/ |
#5
|
|||
|
|||
Gunner wrote:
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. 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 |
#6
|
|||
|
|||
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: On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin wrote: Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? 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. 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 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 .. I never knew any of that stuff...shrug. Grant 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 |
#7
|
|||
|
|||
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: 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 -- __ Pete Snell Royal Military College Kingston Ontario The reasonable man adapts himself to the world; the unreasonable one persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable man. - George Bernard Shaw |
#8
|
|||
|
|||
"Grant Erwin" wrote in message ... Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? snip 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. If so, then balancing isnt going to lower the operating costs anyways..... -- SVL |
#9
|
|||
|
|||
"PrecisionMachinisT" wrote in message ... "Grant Erwin" wrote in message ... Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? snip 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. If so, then balancing isnt going to lower the operating costs anyways..... 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 -- SVL |
#10
|
|||
|
|||
Grant Erwin wrote:
Gunner wrote: 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. 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 You want to measure the current on the meter side of the caps; not on the welder side. -- Keith Bowers - Thomasville, NC |
#11
|
|||
|
|||
"Grant Erwin" wrote in message ... Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? 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. 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 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 |
#12
|
|||
|
|||
Power factor should not affect your electric bill. KWH meters ignore
reactive current. On Tue, 14 Dec 2004 09:46:33 GMT, Gunner wrote: On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin wrote: Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? 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. |
#13
|
|||
|
|||
"Robert Swinney" wrote in message ... "PrecisionMachinisT" wrote in message ... If so, then balancing isnt going to lower the operating costs anyways..... This is true. Power delivered is power sold. 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 |
#14
|
|||
|
|||
On Tue, 14 Dec 2004 14:26:49 GMT, "Pete C." wrote:
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, you might want to check your rate structure. PG&E may have something like a demand charge on its residential (I assume you're a residential customer) rates that kicks in if you have a big surge in demand over something like a 5-minute interval. That sort of thing is not common in residential rates, but PG&E is notorious for its creative rate plans. --RC Gunner wrote: On Mon, 13 Dec 2004 22:29:04 -0800, Grant Erwin wrote: Gunner wrote: Any idea of how to calculate the amount of capacitance needed to reduce idle current in a single phase 220 welder? 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. 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 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 .. I never knew any of that stuff...shrug. Grant 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 Projects expand to fill the clamps available -- plus 20 percent |
#16
|
|||
|
|||
On Tue, 14 Dec 2004 14:07:10 -0800, "PrecisionMachinisT"
wrote: This is true. Power delivered is power sold. 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 ) Only the political stuff you utter is suspect. I read your other posts religiously G Well..and some of the 2nd Am stuff is also suspect VBG Gunner G -- SVL "If I'm going to reach out to the the Democrats then I need a third hand.There's no way I'm letting go of my wallet or my gun while they're around." "Democrat. In the dictionary it's right after demobilize and right before demode` (out of fashion). -Buddy Jordan 2001 |
#17
|
|||
|
|||
If the power bill is painful, have you considered trying to trade for a
gas powered welder/generator? Again if power is that dear, it would likely pay off long term. TMT |
#18
|
|||
|
|||
"Gunner" wrote in message ... On Tue, 14 Dec 2004 14:07:10 -0800, "PrecisionMachinisT" wrote: ( Pretty sure he has me branded as being 'Liberal'--and so anything *I* might say is *definately* suspect ) Only the political stuff you utter is suspect. I read your other posts religiously G "Political stuff" is always suspect, regardless of who it was uttered it. -- SVL " Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others." --Ambrose Bierce |
#19
|
|||
|
|||
Don Foreman wrote:
Power factor should not affect your electric bill. KWH meters ignore reactive current. True for now Don, talking residential that is. But probably not forever here in the USA. New Zealand already charges residential customers for KVARs, and I recall seeing some outfits already selling "whole house" power factor correction boxes here, which IMO are a sham because they don't automatically track the requirement and correct for it. Capacitors of the right size at each load are the easiest solution and also reduce the I^2*R losses in the house's wiring, giving a tiny additional savings there. BTW, there was some work done by some Navy guy a while back showing how much power could be saved by adding a running capacitor to exixting dumb old split phase induction motors. IIRC it increased the efficiency of the average motor by about 10%. Just my .02, Jeff -- Jeffry Wisnia (W1BSV + Brass Rat '57 EE) "As long as there are final exams, there will be prayer in public schools" |
#20
|
|||
|
|||
On Tue, 14 Dec 2004 06:12:36 GMT, Gunner wrote:
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 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 |
#21
|
|||
|
|||
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 On Tue, 14 Dec 2004 06:12:36 GMT, Gunner wrote: 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 |
#22
|
|||
|
|||
On Sun, 09 Jan 2005 23:24:40 -0500, Brian Lawson
wrote: 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 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 On Tue, 14 Dec 2004 06:12:36 GMT, Gunner wrote: 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 "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) -------------------------------------------------------------------------------- |
#23
|
|||
|
|||
Gunner 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. snip Gunner 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 NRA LOH, NRA Life NRA Second Amendment Task Force Charter Founder |
#24
|
|||
|
|||
|
#26
|
|||
|
|||
On 15 Jan 2005 11:40:15 -0800, wrote:
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 Dan, I read it, but its over my head so I simply nodded, and printed it out to try in the near future. Gunner wrote: 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 " 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: 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 "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 |
#27
|
|||
|
|||
|
#28
|
|||
|
|||
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: On 15 Jan 2005 11:40:15 -0800, wrote: Actually, this is starting to **** 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 |
#29
|
|||
|
|||
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 |
#30
|
|||
|
|||
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 |
#31
|
|||
|
|||
http://www.lmphotonics.com/pwrfact.htm 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 |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|