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  • do longer leads=less power

    if i make all my leads 25ft long insted of the original 6ft long will i lose any power

  • #2
    Well, yes, but those distances will be insignificant losses. Don't even think about worrying about it.

    What kind of machine are we talking about? A CC machine (stick/TIG) will output constant amps no matter how long the leads get (within the abilities of the voltage, of course.) As the volts dropped from the longer leads, you would be able to measure power loss, measured in watts, but this is not going to be a useful measurement to a stick welder. The heat is in the amps.

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    • #3
      Originally posted by MAC702
      Well, yes, but those distances will be insignificant losses. Don't even think about worrying about it.

      What kind of machine are we talking about? A CC machine (stick/TIG) will output constant amps no matter how long the leads get (within the abilities of the voltage, of course.) As the volts dropped from the longer leads, you would be able to measure power loss, measured in watts, but this is not going to be a useful measurement to a stick welder. The heat is in the amps.
      The longer leads will cause a I^2R loss down the line.

      The longer the wire, the more resistance the voltage sees and the more voltage drop.

      Since voltage resistance and current are all related from the equation V=IR, the more voltage drop will correspond to a decreased current.

      Power is defined as V*I or I^2R. (W)

      Heat is generated by the combination of the voltage and the current and is not solely dependent on current alone.

      Given the same amount of input power and two different lead lengths, the longer lead length will have less power at the electrode end. The heat loss occurs in the wire. Does it matter over 10 feet? Not unless your leads are under-rated to begin with. Would I worry about it? No. Would I worry about longer lengths - probably. Dependent on the situation.

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      • #4
        Andy,

        Get some #2 welding cable and go for it. Your ArcOne inverter won't notice the difference. If you start talking 100', then you'll want a bigger lead, but you'll be fine at 25'.

        Hank
        ...from the Gadget Garage
        Millermatic 210 w/3035, BWE
        Handler 210 w/DP3035
        TA185TSW
        Victor O/A "J" series, SuperRange

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        • #5
          Since Im a portable welder means that I am always adding a 100' on both sides + and - to my leads that are already 175' long. minus any calculations from the mathmatical wizards we have on this site, it seems to me that I turn my welder up approx. 10 amps for the extra 200' of cable I add to my circuit when I am arc welding, in my estimation the added length in you cable should mean turning up your machine about 1 amp to compensate.
          Just as every one else said ( not enough to worry about ) .

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          • #6
            Originally posted by clintonwelding1
            if i make all my leads 25ft long insted of the original 6ft long will i lose any power
            You will if you use #12..LOL..As Hank said, use some #2. I have #2- 50' each on a 230 amp buzz box and i could push that to 100'
            LICENSED ELECTRICIAN
            BAKERY MECHANIC

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            • #7
              Originally posted by Steved
              Since voltage resistance and current are all related from the equation V=IR, the more voltage drop will correspond to a decreased current.

              Power is defined as V*I or I^2R. (W)

              Heat is generated by the combination of the voltage and the current and is not solely dependent on current alone.

              Given the same amount of input power and two different lead lengths, the longer lead length will have less power at the electrode end. The heat loss occurs in the wire. Does it matter over 10 feet? Not unless your leads are under-rated to begin with. Would I worry about it? No. Would I worry about longer lengths - probably. Dependent on the situation.
              A Stick/TIG machine is a CONSTANT CURRENT power source. Lead length is not significant unless they were SO long that the machine's voltage could no longer drive the current through the resistance of the wire. The heat of an electric arc is related to its current flow, not its voltage.

              House wiring is driven from a CONSTANT VOLTAGE power source (your power company) and some of its normal charicteristics do not apply. Most electricians only think in terms of constant voltage. MIG machines, however, do use constant voltage power sources for the most part.

              I^2*R losses are POWER losses (wattage), calculated from the current and resistance values. A constant current power source, like a Stick/TIG welder, will raise voltage to compensate to keep a constant average current value, therefore:

              Welding heat (temperature of the electric arc, based on current) is not dependent on lead length in a constant current welding machine.

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              • #8
                Mac, that makes good sense as you have laid it out, but I'm curious why it seems like longer leads suck more power, unless they are the right gauge. For example, if you were welding at 240A CC with 400' of 1/0, you'd notice a "performance" diference as oppsed to if you were using 2/0 or 4/0? I've expierenced it, and always atributed it to the increased resistance over that distance (V=IR), but, if the welder raises the voltage then to compensate for the increased R (like you were saying), then something is still causing that difference in apparent performance or "feel"? Now, I've only experienced this with a 15y/o machine, not the new 302. Is there a machine difference that would, at the time, have the older machine respond to the R increase differently than new machines, or is there something I'm just missing?
                hre

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                • #9
                  Originally posted by MAC702
                  A Stick/TIG machine is a CONSTANT CURRENT power source. Lead length is not significant unless they were SO long that the machine's voltage could no longer drive the current through the resistance of the wire. The heat of an electric arc is related to its current flow, not its voltage.

                  House wiring is driven from a CONSTANT VOLTAGE power source (your power company) and some of its normal charicteristics do not apply. Most electricians only think in terms of constant voltage. MIG machines, however, do use constant voltage power sources for the most part.

                  I^2*R losses are POWER losses (wattage), calculated from the current and resistance values. A constant current power source, like a Stick/TIG welder, will raise voltage to compensate to keep a constant average current value, therefore:

                  Welding heat (temperature of the electric arc, based on current) is not dependent on lead length in a constant current welding machine.

                  I really do not want to make this into a pissing match. Just for everyone that is interested:
                  Agreed that power losses are I^R or VI.

                  It is the power delivered where the arc is that generates the heat. To be technical, the arc is generated and sustained by the voltage potential at the tip of the electrode. Not the current. The current is merely the electron transfer. The voltage potential is required to ionize the gas providing an initial lower resistance through the air and resulting in a circuit.

                  At the end of the day, the heat generated is a combination of both the voltage and current. You cannot have one without the other. P=VI=I^2R. To say it is current alone is incorrect.

                  It is not solely dependent on current. It is the current being used x the voltage potential required to cause the current OR/AND it is the current squared x the resistance that the arc sees.

                  In the end, I do agree but do not want people thinking that current causes all of the heat. It does not.

                  Comment


                  • #10
                    Originally posted by Steved
                    the arc is generated and sustained by the voltage potential at the tip of the electrode. Not the current. The current is merely the electron transfer.
                    Agreed 100%. The voltage MAKES the arc. No one is questioning that.

                    There are a LOT of volts (compared to welding) between your finger and your doorknob after scuffing your feet on the carpet. I don't know of anybody who's so much as melted off a fingerprint from this experience.

                    Arclength is another way to vary the resistance in your welding circuit and is far more significant than the difference between 10' and 100' of 2/0 copper. I can weld off my TB with 15' leads and weld with the same settings at 150' with the same arclength and quality.

                    I don't know if cheap tombstone-type machines have the same capabilities to up the voltage when necessary.

                    To summarize: Longer leads have higher resistance. Higher resistance will cause a power loss through voltage drop. Heavier leads will cause less power loss (therefore less heat in the leads.) Lighter leads will get warmer but the full amperage requested still gets through the arc, within the voltage capabilites of the machine.

                    Comment


                    • #11
                      Originally posted by Coalsmoke
                      Mac, that makes good sense as you have laid it out, but I'm curious why it seems like longer leads suck more power, unless they are the right gauge. For example, if you were welding at 240A CC with 400' of 1/0, you'd notice a "performance" diference as oppsed to if you were using 2/0 or 4/0? I've expierenced it, and always atributed it to the increased resistance over that distance (V=IR), but, if the welder raises the voltage then to compensate for the increased R (like you were saying), then something is still causing that difference in apparent performance or "feel"? Now, I've only experienced this with a 15y/o machine, not the new 302. Is there a machine difference that would, at the time, have the older machine respond to the R increase differently than new machines, or is there something I'm just missing?
                      It may be quite common for a lot of machines to be operating at near their maximum voltage abilities at a given current setting. Especially with older machines would there be a smaller range of voltages available at any given current setting. I've used old 3-phase Lincoln motor-generators that allowed you to manually adjust the OCV after the current setting. The label on the dial showed you the voltage range based on the current value set with the other dial.

                      For that matter, it may be that many so-called "constant-current" power sources, really aren't all that "constant" based on the abilities of the machine to vary its voltage to maintain that constant-current when faced with unusually higher resistance values than "stock."

                      A test with a modern, high-quality, professional 3-phase constant-current power source like a Trailblazer 302 with digital ammeter and voltmeter might prove interesting indeed.

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                      • #12
                        So then it seems that lead length and size is a case of how much power you do or do not want to waste, rather than a matter of affecting arc power or quality.

                        FWIW, some might call these things pissing matches, but I enjoy the information that emerges from them when they are kept civilized like this. Thanks.
                        hre

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