16
\$\begingroup\$

I saw on youtube some people performing arc welding with "consumable electrodes". At a first glance, I saw that the current flows through all the electrode and the workpiece and my question comes from this fact.

I think that the workpiece doesn't melt because it is usually a lot bigger than the electrode, therefore, it can dissipate the heat a lot faster. The electrode is thinner though, and I don't understand why the whole electrode doesn't melt if the current that flows through it is high enough to melt the tip of the electrode.

I thought about it and my guess is that it has something to do with the contact resistance at the tip of the electrode being different from that of the electrode's material. The reason is that the power, which is somehow proportional to the heat generated, should be $$P=I^2R$$ But I don't think that the difference between the two resistors is high enough to explain this phenomenon, so I was wondering what part I'm missing!

\$\endgroup\$
  • \$\begingroup\$ If you put 10 Amp thru 2 resistors and one is 0.01 and the other is 1 ohms what is the difference in power? 1W vs 100W? The answer lies in the gas interface arc resistance. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 19 '18 at 13:29
  • 5
    \$\begingroup\$ The workpiece most certainly does melt, although only in the immediate area of the arc - otherwise you end up without a proper weld. A weld requires that the metal of the filler rod and the pieces being welded merge. By definition, unless the workpiece also melts there is no weld. \$\endgroup\$ – UnconditionallyReinstateMonica Apr 19 '18 at 19:34
32
\$\begingroup\$

The electrode's resistance isn't what's heating things up – it's the resistance of the ionized air in the arc!

Hence, things close to the arc get hot, and things farther away don't.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Isn't ionized air a low resitance path? So why does it heat so much? \$\endgroup\$ – Elia Apr 19 '18 at 15:48
  • 13
    \$\begingroup\$ @Elia Compared to unionized air, yes. Compared to metal, it has a much higher resistance. \$\endgroup\$ – Gavin S. Yancey Apr 19 '18 at 16:01
  • 24
    \$\begingroup\$ @g.rocket It took me three reads to realize that "unionized air" means "air that is not ionized", not "air that belongs to a union". \$\endgroup\$ – Jeff Bowman supports Monica Apr 19 '18 at 18:43
  • 19
    \$\begingroup\$ @JeffBowman how to tell a chemist from a electrician, ask them to say "unionized" \$\endgroup\$ – MikeTheLiar Apr 19 '18 at 19:21
3
\$\begingroup\$

When the electrode is brought close to the work piece, the air gap narrows to the point that a spark is created when the electric field strength (in volts per meter, for example) rises high enough to ionize the intervening air molecules.

The ionized air is a plasma, which has a very high temperature - high enough to melt the electrode and the work piece material.

As long as the welder maintains a gap of the right length, the electric field strength will be high enough to ionize the air within the gap and melt the nearby material of the welding rod and the work piece. Some metal may also gasify and the turn to plasma as well, and thus contribute to the arc.

If the gap gets too large, then the plasma will cease, along with any welding.

Anyone who has worked with a stick welder (one that uses welding rods) can tell you that if the gap gets too small, you may touch the rod to the work piece, you may create just enough plasma at the time of contact to weld the rod to the work piece. At that point, you have a continuous metallic circuit with no plasma. It will conduct the same amount of current as it would while doing a proper weld, but, without the plasma arc, nothing will melt.

None of this explanation has anything to do with the resistance of the plasma. It is a function of how the plasma forms in response to the imposed electric field strength.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ The electrode doesn't melt because the current that flows through it is not high enough to melt the tip of the electrode. \$\endgroup\$ – Mazura Apr 20 '18 at 1:10
  • \$\begingroup\$ Since I've never welded I can't confirm it with facts, however, physically speaking does it mean that the the arc won't be created unless the distance 'd' of the electrode from the workpiece is V/d>3k (where V is the voltage of the welder and 3kV/mm the electrical breakdown of air)? For instance, with a voltage of 20V (I read that they use high current and low voltage, so 20V should be reasonable), you would have d<0.0066 mm. Is it humanly possible to keep that distance without touching the workpiece? \$\endgroup\$ – Elia Apr 21 '18 at 8:04
  • \$\begingroup\$ Once you have establish a plasma, you can pull the rod away and still maintain the plasma. So the technique is to bring the rod close enough to spark and then pull back to a working distance. The plasma is conductive. It takes practice to do this with skill. You can see evidence of this by observing how a Jacob's Ladder works. \$\endgroup\$ – Jim Apr 23 '18 at 17:52
2
\$\begingroup\$

There are several welding processes that produce heat through different means. I think TIG welding is conceptually easier to understand than stick or MIG welding. The explanation will help understand other welding processes so I will start explaining TIG welding.

In TIG welding, (gas tungsten arc welding or GTAW), a welding power supply is connected to a hand torch with a tungsten tip. The negative electrode is connected to the torch. The positive electrode is connected to the work piece to be welded.

An arc is created by a circuit in the power supply called an arc starter which produces a high voltage, high frequency pulse between the tungsten tip and the work piece. The arc has enough energy to strip electrons from the shielding gas and create a path of ions that conduct electricity from the tungsten tip to the work piece. For tig welding, argon gas is typically used since its cheap, ionizes easily, and is heavier than air so it keeps oxygen out.

When the ion path is complete, the power supply senses the voltage drop between the electrodes. When there is no ionized path between the electrode and work piece, there may be 50V or more difference between the tungsten and work electrodes. After the arc is initiated, the voltage between the electrodes will drop to around 10V depending on the gap size. At this point, the power supply turns on the welding current. Tig welding is done with constant current power supply.

The arc is maintained by resistive heating of the shielding gas. The ionized gas acts a resistor where the heat is a function of the voltage across the gap and the current through it. The high current through the ionized gas dissipates so much heat that the gas stays hot enough to remain a plasma and continues to conduct.

The heat is not evenly distributed across the arc however. In this configuration that I've just described, the electrons are actually firing out of the tungsten tip and hitting the work piece. This causes the heat to focused on the work piece. If I reversed the polarity of the electrodes and connected the negative to the work piece and the positive to the torch, I would have the opposite effect. I would still get an arc and plenty of heat, but the heat would be focused on the tip not the piece I was trying to weld. This would result would be the tip melting into a ball and falling off. Tungsten is used for the tip since it has the highest melting point of any metal. In tig welding, you do not want the electrode melting and becoming part of the weld but in other types of welding you do.

In MIG welding(gas metal arc welding or GMAW), this is what you want. In MIG welding, the electrode is a conductive wire that is fed from a spool of wire at high speed. The wire melts and becomes part of the weld. The polarity is reversed so that the wire is positive and the work piece is negative. You don't need a arc starter with MIG.

When you squeeze of the trigger of the mig torch, the wire feeder starts pushing out wire. When the wire makes contact with the work, the wire acts as a resistor and heats up. The longer the stickout of the wire, the more resistance it will have and it will produce a different volt drop across it.

Due to the high current through the wire, the wire will melt and burn back. This produces a small gap between the work and wire where there is sufficent voltage to ionize. This creates an arc. Without getting into the specifics of different MIG processes(short circuit, drip, and spray transfer) this process essentially repeats. The wire makes contact. Heats up and melts back. Strikes an arc, then makes contact again. Etc.

|improve this answer|||||
\$\endgroup\$
1
\$\begingroup\$

The workpiece usually needs to melt too (but not too much or you get a material breakthrough), otherwise you would not have a strong mechanical connection. You account for the thickness, thermal mass and thermal conductivity of the workpiece by adjusting the current and material-feed-rate. And as Marcus Müller already said: its not about the electrodes resistance.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.