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Ever see a good welding machine constructed by half-brains? If not, you want to see this! Patient: Welding Machine - 117VAC 24 Amp input, 85Amp AC output.Source: Purchased from Princess Auto for $80 Canadian.Capacity: Welder is rated to use 1/16 inch and 3/32 inch diameter rods.See picture Labot-1:Left top = add for welderLeft bottom = picture of welderRight top = electrical specifications on welder caseRight Bottom = window showing amperage setting 40 to 85 amps. Problem: lack of weld fusion when using Lincoln 7018AC rods of 5/32 inch diameter.Observations: - while making fillet welds on 1/4 thick inch steel plates1 easily possible to sustain too long an arc gap, leading to porous welds with little adhesion2 makes proud welds that tend not to adhere - either the vertical or to the horizontal surface3 welds can be knocked apart fairly easily and reveal partial fusion over only 20% of lengthDiagnosis:1 - open-circuit voltage too high (long arc)2 - welding current too low (proud welds, no fusion)Solution:Examine welder internals and if the transformer configuration permits, remove sufficient turns of wire from the secondary winding of the transformer to reduce secondary output voltage.A limited input power (117 volts x 24 amps = ~2,800 watts) results in a similar value of output power (input power x transformer efficiency: say 2800 watts x 85% = 2,380 watts). The output power (2,380 watts) is the product of output voltage x output current. Thus, decreasing the number of turns on the secondary winding of the transformer will both decrease the output voltage and increase the output current.Exploratory Surgery see picture Labot-2:Left top = the welder opened upRight top = front view primary winding (copper) on left, secondary winding (white) on rightLeft bottom = rear view primary winding (copper) on right, secondary winding (white) on leftRight Bottom = primary winding taps to permit either 115 volt or 230 volt inputCurrent adjustment turning a knob on the front panel drives a section of steel laminations in or out of the transformer core. When out, the transformer functions with maximum magnetic coupling between the primary and secondary windings and maximum current output is obtained. When the laminations are driven in, the magnetic coupling between the primary and secondary decreases and the output current limit drops. Picture Labot-3 shows the laminations being drive in.Left top = steel laminations out of transformer core maximum welding currentRight top = steel laminations driven about half way into transformer coreLeft bottom = steel laminations driven all the way into transformer core minimum welding currentTransformer Construction60 cycle AC transformers are constructed of thin laminated steel sheets, each insulated from each other. Why? This stop major eddy currents from flowing in the transformer metal and uselessly burning up power heating up the transformer.See picture Labot-4.Left top = Note the weld across the top of the transformer laminations! IDIOTS!That weld just ruined the transformer!Right top = A check with a clamp-on ammeter shows 5.2 amps going into the welder at idle.Left bottom = The back of the fan says it draws only 0.25 amps. That means that about 5 amps is going into heating the darn transformer under no load!! Hey thats 115 volts x 5 amps = 575 watts! Thats 20% of our input power (2800 watts) just blown away! We could have turned that 575 watts into about 20 amps of much needed welding current idiots! How dumb can you get ????Anyway
As the photos show, the transformer was of a configuration suitable for modification the primary (117VAC) winding and the secondary winding (welding output) were both visible and accessible.I was able to count the number of layers of wire and the turns per layer.Primary Configuration: Copper wire ~12 or 14 gauge: 18 layers of 14 turns each = 252 turns.Secondary Configuration: Aluminum wire ~6 gauge: 9 layers of 8 turns each = 72 turns.Transformer turns ratio = Primary Turns / Secondary Turns = 252 / 72 = 3.5With a 117 volt input, the secondary output voltage will be 117 volts / 3.5 = 33.4 voltsWith a 24 amp input, the secondary output current should be 24 x 3.5 = 84 amps**However, allowing for transformer efficiency (85%), the secondary output current could be more like 84 amps x 85% = ~72 amps.So, you can figure out what to expect with a change in the number of turns on the secondary.Secondary Volts per turn = 33.4 volts / 72 turns = 0.464 volts per turnSecondary Amps per turn = 72 amps / 72 turns = 1.0 amps per turn (at 85% efficiency)OrSecondary Amps per turn = 84 amps / 72 turns = 1.167 amps per turn (at 100% efficiency)Lets put that in a table to see what our options are.Table of Secondary Winding Estimated Effect of Turns on Voltage and Current Transformer Efficiency 85% 100%Turns Voltage Amps Amps72 33.4 72 8471 32.9 73 8570 32.5 74 8669 32.0 75 8868 31.5 76 8967 31.1 77 9066 30.6 78 9165 30.2 79 9264 29.7 80 9363 29.2 81 9562 28.8 82 9661 28.3 83 9760 27.8 84 9859 27.4 85 9958 26.9 86 10057 26.4 87 10156 26.0 88 10355 25.5 89 10454 25.1 90 10553 24.6 91 10652 24.1 92 10751 23.7 93 10850 23.2 94 110At first, I removed 8 turns (72 8) = 64 turns left and tried welding with that. It was better but I still didnt have enough amps to made a good weld.See picture Labot-5 for wire coming off.I then removed another 7 turns (64 7) = 57 turns left and tried again. This was a marked improvement see left bottom photo.What was the welding current? I used a clamp-on AC ammeter to measure the AC current through the ground lead leading to the work; I measured ~70 amps welding and 115 amps if I stuck a rod. Those numbers are way, way below the supposed rating of this welder!!! This welder, as delivered, was not capable of producing anywhere near the claimed 85 amps output!! What a mess! Welds across the transformer, and way too many turns on the secondary who the building such CRAP??? Now the quality of the components was pretty good - but what's going on during assembly??? Net Result = POOR!Anyway, I got it working not too bad and it makes a nice weld!I am not about to pull that darn transformer all apart to gain another free 20 amps
what a waste!!!! Somebody sure dropped the ball making these welders. Anybody say, "Quality Control" ??? Folks, I discovered that my welder had a previous lobotomy before I attempted mine! Just Amazing!!!Rick V Attached Images
Reply:Amazing that you got anything for an 80 dollar 110 V welder. When I buy something with that "powerfist" label my expectations are limited.
Reply:With that kind of engineering knowledge and ability, I'm going to bet that you can afford to buy yourself an arc welder that actually works right from the start!Very impressive job you did there.
Reply:Couple notes:the 5amps you read with the clamp on meter was primarily reactive current, not loss to heating. A little power is lost in the resistance of the windings and in core hysteresis, but not much.The weld running across the laminations hurts nothing. It won't provide a significant path for eddy current losses.Open circuit voltage too high? Doesn't matter. YOU set the arc voltage by controlling the arc length when welding. Higher OCV is generally a good thing with these welders. Makes it easier to strike and maintain an arc. It does effect the C-V curve (higher OCV is generally a steeper curve) and arc characteristic.The movable core is called a shunt.If you are trying to run 5/32 7018AC on this machine, you will have no luck. 3/32 should be doable at the top end-- 70 to 75A is sufficient, but at the cold end. Did you mean 3/32?
Reply:Hi enlpck - I'll answer your questions.RE: the 5amps you read with the clamp on meter was primarily reactive current, not loss to heating.Hmmm, you could be partially correct; would it make any difference if I remeasure using a resistor in line and measuring the voltage drop across it?RE: A little power is lost in the resistance of the windings and in core hysteresis, but not much.No - this transformer does heat up a lot (very hot to touch after say 10 minutes) with no load. That would not happen if there were not considerable eddy current heating.Also, I have considerable past experience with eddy current heating of a transformer.e.g. I was making a battery charger. The transformer laminations were held together with 4 steel screws (each with a non-conducting cover sleeve) passing through (across) the transformer laminations. Without a load, the transformer did not even get warm to the touch. However, when I added on an aluminum heat sink, I needed long screws - got em, put em in but the insulating sleeves were not long enough ... the transformer got quite HOT! I had to put heat-shrink tubing over the four screws and use nylon washers and nuts. The transformer then ran cold as before. A few years later, the heat-shrink sleeves shrunk - and wow - it ran 'HOT' again. I fixed the insulation - ran cold again.RE: The weld running across the laminations hurts nothing. It won't provide a significant path for eddy current losses.I don't agree - based upon my experience related above.A weld across the transformer in the welder will definetly heat the transformer up... and that is what is happening - the transformer gets very hot to touch just idling - not even used for welding yet.RE: Open circuit voltage too high? Doesn't matter. YOU set the arc voltage by controlling the arc length when welding. Higher OCV is generally a good thing with these welders. Makes it easier to strike and maintain an arc. It does effect the C-V curve (higher OCV is generally a steeper curve) and arc characteristic.I have hands-on experience with this particular machine. Originally, I could draw a very long arc (that is not open circuit) while welding. With less secondary turns I no longer can produce a long arc like that. I find the higher amperage available is more an advantage (rods don't stick) than the previous higher output voltage. RE: If you are trying to run 5/32 7018AC on this machine, you will have no luck. 3/32 should be doable at the top end-- 70 to 75A is sufficient, but at the cold end. Did you mean 3/32?Yes - you got me there; I meant 3/32.It is important to get away from 1/16 or 5/64 rods because of expense ($5/pound) and limited rod selection (typically 6013 only). At 3/32, a whole new world opens - $2/pound and 6011, 7014 and 7018, etc. is available.Thanks,Rick V
Reply:Hi Folks! After enipck insightful comments, I looked into everything more carefully. He was correct that eddy current losses under no load were nowhere near as high as I calculated by 115 volts x 5 amps = 575 watts!; loss was more like 25 watts! I got it wrong - 1st time experience!The detailed calculations for inductance reactance, other numerous factors and the confirming measurements that I took are covered in 12 pages of notes shown in Labot-6 ...and Labot 7(If you really want this stuff in detail, send me a private message with your email address; Ill email you higher resolution images of my notes.)I need to mention a few things see images in Labot-81 The BK Precision 875B inductance meter readings were quite misleading (low) for measuring the inductance of the transformer primary coil. I had to use another method (very simple and cheap) described by Barry Hansen at url http://www.oz.net/~coilgun/theory/measureinductance.htmExamples:The BK meter readings of the primary coil inductance versus the simple method were:Open circuit secondary = 17.3 mH (BK)versus 46.4 mH (simple)Short circuit secondary = 16.4 mH (BK)versus 31.8 mH (simple)2 The PowerFist DT266 clamp on AC ammeter gives different readings depending on where the wire was inside the clamp especially when the currents are low often the case here. In such cases, I measured current by voltage drop across a resistor.3 I was stymied as to how to measure power loss due to eddy currents in the transformer laminations. I came up with the idea of measuring the heat generated by the welder at idle and comparing that to the heat generated by incandescent light bulbs (40, 60 and 100 watt). So, I put the welder in an insulated cooler. Using an outdoor thermometer, I measured the temperature rise and fall over time caused by operating the welder for fixed periods of time under no load. See image Labot-8.The welder (with fan on) generated the same time-temperature curve as a 60 watt light bulb (welder fan on). Subtracting off the power used by the fan (31 watts) and the I^2R loss from the primary wire resistance (3.6 watts), eddy current heating = ~25 watts.Summary of FindingsPrimary Coil Measurements:230 turns of #10 gauge copper wireResistance = 0.205 ohmsXl Impedance (no load) = 17.5 ohmsXl Impedance (shorted secondary) = 12.0 ohmsInductance (no load) = 46.4 mHInductance (shorted secondary) = 31.8 mHSecondary Coil Measurements:Was 72 turns of equivalent to #6.5 gauge aluminum wireNow 57 turns of equivalent to #6.5 gauge aluminum wireResistance = 0.0446 ohmsOPEN CIRCUIT SECONDARY (TRANSFOMER ON BUT AT IDLE)Primary Coil losses:Input volts = 115, apparent current = 4.0 ampsApparent input power = 115 X 4.0 = 460 wattsPower loss to fan = 31 wattsPower loss to primary wire resistance = 3.6 wattsPower lost to eddy currents in XMFR lamination = 25 watts Total primary coil losses (~measured) at idle = ~60 watts.SHORT CIRCUIT SECONDARY (TRANSFOMER ON DELIVERING MAX POWER)Measured secondary output current = 115 ampsPrimary Coil losses:Input volts = 115, apparent current = 34.3 ampsApparent input power = 115 X 34.3 = 3,944 wattsPower loss to fan = 31 watts (same as at idle)Power loss to primary wire resistance = 242 wattsPower lost to eddy currents in XMFR lamination = 221 watts Total Predicted power loss at shorted output = 494 wattsSecondary Coil losses:Power loss to secondary wire resistance = 590 wattsTotal Predicted power loss at shorted output =Primary losses + Secondary losses = 494 + 590 = 1,084 wattsThus output power = input power losses = 3,944 1,084 = 2,860Welder Efficiency = Output/Input power = 2,860/3944 = 72.5%Observations on Power LossesAt full output, the big power eaters are from:a) Secondary wire resistance heating = 590 wattsb) Primary wire resistance heating = 242 wattsc) Power lost to eddy currents in XMFR lamination = 221 watts Can we do anything about power losses due to a), b) or c)?a) Secondary wire resistance heating = 590 wattsThe biggest power eater in the secondary winding mage of #6.5 gauge aluminum wire. Changing to #6 copper wire (only 6% more volume) would improve things, dropping only 318 watts. This would free up (590 318) = 272 watts. At 25 volts output, that would give an additional 272/25 = 10.8 amps of welding current.Actually, there is room to use #5 gauge wire, which would give 13 amps more output.b) Primary wire resistance heating = 242 wattsThe primary winding is good #10 copper wire. There might (???) be space for #9 wire, if so that would release 122 watts at 25 volts output about 4.5 amps of welding current. However, changing out the primary would be a major a pain in the ask-me-no questions.c) Power lost to eddy currents in XMFR lamination = 221 watts Here is where the use of non-insulated steel pins and no welds across the transformer laminations would have paid off. Based upon my earlier experience with another transformer, I would expect at least a ten fold reduction in eddy currents. Better construction technique could have made available about 200 watts of free power. At 25 volts output, this means about 8 more amps of welding current.If transformer construction, primary and secondary windings had been optimized during manufacture, it looks like there would have been an additional 25 amps (13 + 4.5 + 8) of welding current available and welder efficiency would have risen from 72.5% to 87.5%.e.g.Primary losses + Secondary losses = 172 + 318 = 490 wattsThus output power = input power losses = 3,944 490 = 3,454Welder Efficiency = Output/Input power = 3,454/3944 = 87.5%Can I do anything myself? About the only thing I could do (still a big pain) would be to change out the secondary winding from the present aluminum wire to #6 copper wire (I have that in my junk) and gain about another 10 amps of welding current.Comments or suggestions?Rick V Attached Images
Reply:You have way too much time on your hands.. ...zap!I am not completely insane..Some parts are missing Professional Driver on a closed course....Do not attempt.Just because I'm a dumbass don't mean that you can be too.So DON'T try any of this **** l do at home.
Reply:Gotta agree...and it´s great that you can show us how simple stuff can fix such a mess, or a "80 bucks" welder....Sent you a PM, we looking at eddy currents in physics...would be pretty nice if I saw your calculations closer Thanks for the detailed explanation!My Babies: HF Drill pressHF Pipe Bender3 4.5" Black and Decker angle grindersLincoln Electric PROMIG 175that´s it!
Reply:You have way too much time on your hands..Zapster
Reply:Originally Posted by Rick V...different aspects of the trade/technology appeal to different persons; me, I like improving capability. Somehow, I seem on a track destined to turn welding concepts/things that almost work into things that actually do work...
Reply:Sent you a PM, we looking at eddy currents in physics...would be pretty nice if I saw your calculations closer - elvergon
Reply:On the Rod Measurements. . .After unwinding 15 turns from the transformer secondary, I was left with ~56 turns of wire on the secondary coil of the transformer (230 turns on the primary).I measured output voltage and current and saw:- 0 amps at 46.8 volts (open circuit, welder idling), - 85 amps at 23.5 volts while welding, and- 115 amps at 2.2 volts with a stuck (shorted) rod.However, I did not have values for input voltage and current.So, I repeated the exercise and measured both primary and secondary coil voltage and current at the same time.Note: I obtained significantly higher welding currents this time due to better grounding of the work piece = IMPORTANCE OF SOLID GROUNDING! Primary volts and amps were measured at the transformer (bypassing) the fan. I used a 2 foot #10 gauge extension of one primary wire to get my clamp-on AC current meter away from the magnetic field in the vicinity of the transformer.Secondary volts and amps: Volts were measured between the cold end of the rod (in the rod holder) and the work piece. Amps were measured using a clamp-on AC current meter around the ground cable going to the work piece current meter was about 2 feet away from the transformer.A) Open circuit, welder idling- Primary 121.5 volts, 4.5 amps*, Secondary 47.0 volts, 0 ampsB) Welding 3/32 Lincoln 7018AC- Primary 116.3 volts, 40.2 amps*, Secondary 22.8 volts, 102 ampsC) Stuck rod (short circuit)- Primary 116.7 volts, 51.8 amps*, Secondary 1.8 volts, 130 ampsRough Efficiency calculation:Input: Watts = volts x amps = 116 x 40 = 4,640 wattsOutput: Watts = volts x amps = 22.8 x 102 = 2,326Efficiency = Output/Input = 2,326 / 4,640 = ~50% Lost Power = 4,640 watts 2,326 = 2,314 wattsWhere did those 2,314 watts go? Primary wire resistance measured as R = 0.205 ohmsAt 40 amps, I^2R loss = 40 x 40 x .205 = 328 wattsSecondary wire resistance = 0.045 ohmsAt 102 amps, I^2R loss = 102 x 102 x 0.045 = 468 wattsWelding cables total 12 feet of #8 gauge, R = 0.0077 ohmsAt 102 amps, I^2R loss = 102 x 102 x 0.0077 = 80 wattsTotal I^2R (heat) losses = 328 + 468 + 80 = 876 wattsThis leaves (2,314 876) = 1,438 watts unaccounted for.I did a theory calculation of eddy current losses (dont know how accurate that was) but that would only be ~250 watts.That still leaves (1,438 250 watts) = 1,188 watts unaccounted for.Anybody know where those 1,188 watts went?Do they / did they even exist???? *Suspicious Current Behavior! The 30 amp circuit breaker at the electrical panel never popped, neither while welding with a supposed 116 volt input at 40 amps nor during a short circuit with a supposed 51.8 amps.Is the clamp-on AC current meter accurate for reading inductive loads? Note: The clamp-on AC current meter gave the same welder idle current (~4.5 amps) into the transformer primary coil as that measured using the voltage drop across a series resistor inserted in the circuit. So, the clamp-on meter is accurate.However, we know that welder idle current of 4.5 amps (no matter how I measured it) was misleading since the total heating power into the welder amounted to only 60 watts (as measured by temperature rise in the cooler). 31 watts was used by the fan, 3.6 watts I^2R loss was from the resistance of the primary wire, and the rest ~25.5 watts was attributed to eddy current heating of the transformer. Now 60 watts at 115 volts is only a current of 0.52 amps, not the 4.5 amps that I measured. This makes me wonder about the validity of the AC current measurements especially for the primary where the load is inductive! If the real input current is lower, that would make the input power lower and make the efficiency of the welder higher than 50%. Anyone got any suggestions to offer?Rick V
Reply:Gotta love the inquisitive engineer mind.edited for content - Another member complained, not sure if the statement was meant to be humerous or serious. MICROZONELast edited by MicroZone; 05-09-2007 at 12:01 AM.
Reply:Originally Posted by Rick VOn the Rod Measurements. . .Anyone got any suggestions to offer?Rick V
Reply:Anybody know where those 1,188 watts went?
Reply:I would guess that much of the "missing" energy was never there. The current measured on the input will be the vector sum of the real current and the reactive current. The reactive current is 90 deg out of phase with the real current, and reactive current at the input will not be seen at the output. The data given shows a power factor of about 0.75, which is in the ballpark of what I would expect.
Reply:Ricky V,No need to berate Zap for all that he does...he has helped many people make welds that will protect consumers from "failure" (please read as harm). Pardon my French, but from our perspective "playing around" with $40 arc welders, a product of high inherent variation, while manufacturing products for the average customer/consumer is not putting your best foot forward. The "what" we do and "how" we do it is what drives our business margins...not what we can get away with in respect to non-repetitive success. No matter what theoretical mathematics say the outcome of your product will have inherent variation due to the material that you started with, and unfortunately that variation may be experienced by the end customer/consumer. As a hobbyist I would see no fault in your experiments. Please understand that I have no desire to berate you in your efforts to continuously improve upon an existing design. We all have a duty to point out that saving money is desirable, however, when it is done in a setting where a customer or consumer may be detrimentally affected, using tools and components that are substandard, ethically we must react. Saving monies at the expense of any customer/consumer only promotes continuous global outsourcing of our core competencies...understanding metallurgical principles and production of reliable fusion between metals in critical areas. Please do not mis-interpret me because I do appreciate your cerebral approach, but I ask that you don't berate those that selflessly mentor others. I plan on posting no further responses to this thread, and will respectfully go back to my livelihood performing sanitary work and continuous education for the sake of my apprentices. I am just a welder who wants to perform work that I am proud of and if I can educate others in the trade as I get ready to retire I feel even better. Take care and feel free to keep questioning the status quo.....
Reply:By the way I'm kidding about berating Zapster and your calculations are outstanding!!!!!!!!!!!!!!! Nice job!
Reply:enlpck - I would guess that much of the "missing" energy was never there. The current measured on the input will be the vector sum of the real current and the reactive current. The reactive current is 90 deg out of phase with the real current, and reactive current at the input will not be seen at the output. The data given shows a power factor of about 0.75, which is in the ballpark of what I would expect.
Reply:CALCULATING POWER FACTOR AND EFFICIENCY OF MY AC STICK WELDERIf you don't like Math - just skip down to the FINDINGS. PRIMARY COIL MEASUREMENTS230 turns (~253 feet) of #10 gauge copper wireResistance = 0.20526 ohmsSECONDARY COIL MEASUREMENTSWas 72 turns of equivalent to #6.5 gauge aluminum wireNow 56 turns of equivalent to #6.5 gauge aluminum wireResistance = 0.0446 ohmsWelding cable resistance = 0.00769 ohms (12 ft of #8 gauge cable)MEASUREMENTS WITH TRANSFOMER ON BUT AT IDLEOutput of secondary coil voltage = 47 volts, current = 0Input to welder: voltage = 121.1 volts, current = 4.5 ampsMEASUREMENTS DURING WELDINGInput to the welder:Input volts = 116.3, input current = 40.2 ampsOutput from the welder:Average volts between rod cold end and work piece = 22.8 voltsAverage output amps = 102 ampsMEASUREMENTS & CALCULATIONS WITH TRANSFORMER AT IDLEApparent input power = 121.5 volts X 4.5 amps = 546.75 volt-ampsPower loss to fan = 30.9 wattsPower loss to primary wire resistance = I^2 x R = 4.5^2 amps x 0.20526 ohms = 4.15 wattsTotal welder power losses (by temperature measurement) = ~60 watts.Power lost to hysterises and eddy currents in transformer = (Total welder power used fan power loss primary wire power loss)= 60 watts 30.9 watts 4.15 watts = ~25 wattsThus, the eddy current power loss is ~25 watts with 4.5 amps flowing in primary current.CALCULATIONS DURING WELDINGPower input to the welder = 116.3 volts x 40.2 amps = 4675.26 VAImpedance Z = V / I = 116.3 volts / 40.2 amps = 2.89303 ohmsPower output from welder = 22.8 volts x 102 amps = 2325.6 wattsPower used inside welderFan = 30.9 wattsPower loss in primary wire resistance = I^2 x R = 40.2^2 amps x 0.20526 ohms = 331.71 wattsPower loss in secondary wire = I^2 x R = 102^2 x 0.0446 ohms = 464.02 wattsPower loss in secondary welding cables = I^2 x R = 102^2 x 0.00769 ohms = 80 watts.Lets assume that hysterises and eddy current loses in the transformer are linear with the magnetic field which is linear with the primary current. Then transformer losses = 40.2 amps(welding) / 4.5 amps(idle) x 25 watts(idle) = ~ 223 wattsSum of all power loses = 30.9 + 331.71 + 464.02 + 80 + 223 = 1129.63 watts.The power used inside the welder and the power used as output by the welder totals 1129.63(inside) + 2325.6(output) = 3455.23 watts. We imagine that all the power used is caused by a single resistor of value R in the primary circuit. Watts = I^2 x R, thus R = watts / I^2 = 3455.23 / 40.2^2 = 3455.23 / 1616.04 = 2.13808 ohms.Knowing the impedance Z (2.89303) and R (2.13808), we can find the inductive reactance XL of the primary coil while welding. From Z^2 = R^2 + XL^2 we have XL^2 = Z^2 - R^2XL^2 = 2.89303^2 2.13808^2 = 3.79824 and XL = 1.94891True Power P = I^2 x R = 40.2^2 x 2.13808 = 3455.22 wattsReactive Power Q = I^2 x XL = 40.2^2 x 1.94891 = 3149.51 VARApparent Power S = I^2 x Z = 40.2^2 x 2.89303 = 4675.25 volt-ampsPower Factor = True Power / Apparent Power = 3455.22 / 4675.25 = 0.739From cos(phase angle) = 0.739, the phase angle = 42.35 degrees.FINDINGSSo what have we got?This welder needs 4675 watts to operate. However, only 3455 watts (~74%) does anything. The difference (4675 3455), 1220 volt-amps reactive moves back and forth making things happen but doesnt actually consume any power. Of the remaining 3455 working watts, 1129 watts heats up the welder internals and only 2325 watts (50% of the 4675 watts needed to operate) is actually available for welding! Only 1/2 the input power actual goes into making the weld!I aint kidding folks this is the way it is. Most transformer based welders will be quite similar in behaviour.The only practical way to beat this phenomena is to use an inverter-design welder that can increase the Power Factor towards 1 then you convert most all your input power draw into output amps.Take a look at a commercial inverter welder operating as a stick welder.Lincoln Electric Invertec V205-TInput Power = 115 volts, 34.0 amps, Output Power = 24.4 volts, 110 amps @ 35%Output watts / Input watts = 24.4 x 110 / 115 x 34 = 2684/3910 = 68.6%Input Power = 115 volts, 26.0 amps, Output Power = 23.6 volts, 90 amps @ 60%Output watts / Input watts = 23.6 x 90 / 115 x 26 = 2124/2990 = 71.0%[U]So a good inverter welder can operate at ~70% efficiency with DC output.[/U]My transformer-based AC only welder operates at 50% efficiency.If I had to convert from AC to DC, there would be significant additional power loses in the rectifiers and another Power Factor reduction from an additional inductor the large choke coil that would be required to stabilize the arc. [I]We can compare commercial TIG welders and see something similar.[/I]Example: From the 2006 Lincoln Product Catalog:Transformer: Precision TIG 185Input Power = 230 volts, 64 amps, Output Power = 27 volts, 185 amps @ 15%Output watts / Input watts = 27 x 185 / 230 x 64 = 4995/14720 = 33.9%Input Power = 230 volts, 64 amps?, Output Power = 24 volts, 90 amps @ 100%Output watts / Input watts = 24 x 90 / 230 x 64 = 2160/14720 = 14.6%Inverter: Invertec V205-TInput Power = 230 volts, 28.8 amps, Output Power = 18 volts, 200 amps @ 40%Output watts / Input watts = 18 x 200 / 230 x 28.8 = 3600/6624 = 54.3%Input Power = 230 volts, 14.2 amps, Output Power = 15.6 volts, 140 amps @ 100%Output watts / Input watts = 15.6 x 140 / 230 x 14.2 = 2184/3266 = 66.8%Jeez, at the end of the day, we appear to be lucky indeed to get half the power out that we started with! Rick V
Reply:Your machine, based on your numbers, is operating at about 66% efficiency. The 'reactive' power doesn't count. It DOES lead to a small bit of loss heating the wires and circuit breakers, and needs to be considered sizing the power feed, but it IS NOT power delivered or used.You can move the power factor toward 1 with a capactior, quite readily.
Reply:Hey enlpck,Your previous suggestion that "I would guess that much of the "missing" energy was never there" proved correct - the missing 1,188 watts turned out to be reactive power. Your prediction of a "power factor of about 0.75" was right on - I calculated 0.739. Your machine, based on your numbers, is operating at about 66% efficiency. The 'reactive' power doesn't count. It DOES lead to a small bit of loss heating the wires and circuit breakers, and needs to be considered sizing the power feed, but it IS NOT power delivered or used
Reply:Originally Posted by Rick VYes and No. Yes, if you are getting your 66% efficiency number from Welding Power / True Power = 2325 watts / 3455 watts = ~67%.No, in that the Power Company will measure and charge for Apparent Power (4675 volt-amps). The True Power is 3455 watts with 2325 of that going into actual welding.
Reply:Also... power factor correction is done all the time. The caps arn't really that big, but they arn't cheap (though cheaper than larger wire and a larger breaker when running 100ft for a synchrowave 350). On many machines, the capacitor is on a transformer winding rather than directly across the input leads. The can reduce the required capacitance, at the expense of increasing the voltage rating needed. A service benefit is a reduction in the primary winding reactive current.Normally, one doesn't correct any more than needed. Bringing the PF to 85% or better satisfies some power companies, and 90% will satisfy most of them, in terms of the supplier looking for the consumer to remediate the poor power factor.
Reply:You guys ought to be working on the space program.No. The power company does not charge for volt-amps. The real power is measured and charged for.
Reply:Originally Posted by Rick VOk, that is good news.Hey enlpck, have I got it right in that - there is still extra current (reactive: going to and fro) that will affect supply cable size and breaker size? Will power factor affect the current sensed by the circuit breaker in my electrical panel?E.g. does this mean for example that I might need a 40 amp breaker and 10 gauge wire for a power factor of 0.75 instead of a 30 amp breaker and 12 gauge for a power factor of 1?
Reply:Conversion from 120 VAC to 240 VACI substantially increased the output current of this welder from ~80 amps to 110 amps.I did this by reducing the number of turns of the transformer secondary winding from 76 to 52 turns. However, the input current to the transformer primary also rose, from ~32 amps to 40 amps.To handle the original draw of 32 amps, I had installed a special circuit in my garage using #10 copper wire. For use outside my house, I had made special Y-adapter to allow me to use the combined outputs of two normal 15 amp outlet boxes of a regular house.But 40 amps is a lot of current for a 120 volt circuit. So, maybe it is time to convert the welder to operate on 240 volts? Then the current required would drop to 20 amps. In my garage I already have a 240 volt welder outlet and outside my house I could use the dryer outlets of regular houses.When examining the inside of this welder, I noted some things that might have to be changed for operation on 240 volts:1 Transformer,2 Cooling fan3 Off/On light in the off/on switch4 Over-heat lamp 1 - TransformerI had noticed that the transformer had two primary winding instead of one. See Figure 1. This welder is made in China for mass export to many countries: to North America where 120 volt outlets are common and to the rest of the world where 240 volts outlets are standard. Thus by using 2 windings, the manufacturer can supply both markets: putting windings in parallel for 120 volt input or putting windings in series for 240 volt input.Can I simply change the windings from parallel on 120 volts to series on 240 volts?Simple answer YES! MATH if you want it
. else skip to END OF MATH. The measured inductance (L) of the windings in parallel is 48.3 milliHenries (mH) for an inductance reactance of XL = 2 x pi x 60 Hz x L = 18.2 ohms.When the windings are separated one might expect to find that the inductive reactance of each winding would be double, about 36 ohms, since inductors (coils) in parallel behave just like resistors in parallel. E.g. Two 36 ohm resistors in parallel act like one 18 ohms resistor.However, that is not the case here! Why? Here, the windings (coils) are not really separate as both windings are wound on the same form and the mutual inductance of one is influenced by the other. In fact, when the windings were separated, the inductance between 1 and 2 was measured as 49.6 mH and between 3 and 4 was 48.8 mH.When we put the windings in series, we will be doubling the number of turns of the coil and we can expect the inductance to increase by 4 - approximately the square of the number of turns; from ~49 mH to 4 x 49 mH = 196 mH. The actual measured value of the windings in series was 193.7 mH. This is an inductive reactance of XL = 2 x pi x 60Hz x 193.7 mH = 73 ohms.With 120 volts input and the windings in parallel we had XL = 18.2 ohms. This gave a primary winding current of I = V/XL = 120 / 18.2 = 6.59 amps and a power P = V x I = 120 x 6.59 = 791 watts.With 240 volts input and windings in series we have XL = 73 ohms. This gives a primary winding current of I = V/XL = 240 / 73 = 3.29 amps and a power P = V x I = 240 x 3.29 = 789 watts.Notice the no-load power is almost exactly the same for the windings in parallel on 120 volts or in series on 240 volts. This means that we can in fact operate the transformer either way for welding: using 120 volts and 40 amps to supply two windings in parallel or using 240 volts and 20 amps to supply two windings in series.END OF MATH2 Cooling fanThis was a problem as there were no double windings as with the transformer so I couldnt simply reconfigure the wiring. The fan is 30 watts, drawing about 0.25 amps at 120 volts.Change the fan? A quick look about in the electronic and surplus stores showed that in North America this is not an easy item to find a 4.5 inch diameter cooling fan operating on 240 volts.Add a series resistor?I could have put a 30 watt resistor in series with the fan and drop 120 volts across the resistor and 120 volts across the fan. Resistor value = V / I = 120 / 0.25 = 480 ohms. A quick look about in the surplus stores showed this was not an easy item to find: 480 ohms or so is no problem but at 30 watts??? no luck.Operate the fan off a separate 120 volt transformer?Yes on ebay I found lots of foreign travel adapters (240 to 120 volt) for cheap the most suitable type being the 50 watt autotransformer type as shown in Figure 2.Figure 3 shows the wiring of an autotransformer - the primary and secondary share the same winding so there is no isolation between windings. Still, it worked fine for this application.I purchased one 50 watt foreign travel adapters, took out the transformer and installed in my welder to operate the cooling fan worked like a charm! 3 Off/On light in the off/on switchHey, I lucked out; the light was a neon bulb with small current limiting resistor. I noted that the switch body said it was rated for 250 volt operation no mention of 120 volts. Maybe I wouldnt have to change anything? So, I powered the switch on 240 volts no problem, the lamp lit up a mite brighter but the resistor was sized to operate the neon lamp at 240 volts! 4 Over-heat lamp Also a neon bulb rated for 250 volt operation! Perfect, I didnt have to change anything! So, I was able to convert the welder from 120 volt input to 240 volt input by a simple modification to the transformer primary wiring and by adding a small 50 watt, 240 to 120 volt autotransformer to operate the cooling fan. Oh yeah, of course I also had to change the power input plug to a 240 volt welding plug. Happy Camper.Rick V Attached Images
Reply:Originally Posted by Rick VOperate the fan off a separate 120 volt transformer?Yes – on ebay I found lots of foreign travel adapters (240 to 120 volt) for cheap – the most suitable type being the 50 watt autotransformer type as shown in Figure 2....I purchased one 50 watt foreign travel adapters, took out the transformer and installed in my welder to operate the cooling fan – worked like a charm!
Reply:Brilliant...Ha! enlpck, I wish I had thought of that!!! Man... just when you think you outsmarted the system,somebody says, "hey, why didn't you just..."So, I found out... I am not that smart after all!That's Great!FUNNY!!! Thanks enlpck!Rick V
Reply:Hey, it's not as bad as I thought!In the original wiring, the power to the transformer (but not the fan) goes through a heat-sensitive breaker. If the transformer gets too hot, the breaker opens - stopping current flow throught the transformer, but the fan continues to operate - cooling the transformer.. If I wired the fan off the main welding transformer as enlpck suggested, I would lose power to the fan and be left with a 'hot' transformer with no cooling fan. By using a separate small transformer to power the fan, I retained the original safety concept - if the welding transformer overheats and current through the transformer is cut, the fan continues to run and cools the transformer.Rick V
Reply:Ummm....so transformy no worky so good?Nice read - but my head hurts.....
Reply:[QUOTE=Rick V]Hey enlpck,Hmmm, I don' think so - not practically because the capacitor would be huge - not something you are going to find easily. Just to be a bonehead....I had three 1-farad caps in my truck for my sound system. They make a huge difference. hehehehNo, not micro farads, I had three 1-farad caps.Lincoln Power Mig 210MP MIGLincoln Power Mig 350MP - MIG and Push-PullLincoln TIG 300-300Lincoln Hobby-Weld 110v Thanks JLAMESCK TIG TORCH, gas diffuser, pyrex cupThermal Dynamics Cutmaster 101My brain
Reply:[QUOTE=Joker11] Originally Posted by Rick VHey enlpck,Hmmm, I don' think so - not practically because the capacitor would be huge - not something you are going to find easily. Just to be a bonehead....I had three 1-farad caps in my truck for my sound system. They make a huge difference. hehehehNo, not micro farads, I had three 1-farad caps.
Reply:Hi folks, The topic appears to be adding capacitors (capacitive reactance) in parallel with the transformer primary coil in order to reduce its inductive reactance. Recall that "Even for 85% correction (Power Factor) I would need about 1156 microfarads".Joker11 - Just to be a bonehead....I had three 1-farad caps in my truck for my sound system. They make a huge difference. |
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发表于 2021-9-1 00:09:23