He + Ag mix for steel?

admin

I know that this mix works for larger alum and tig, but does it work the same for steel and stainless?G
Reply:Originally Posted by gordfraserI know that this mix works for larger alum and tig, but does it work the same for steel and stainless?
Reply:Helium and Silver?Who is John Galt?
Reply:Originally Posted by bassboy1Helium and Silver?
Reply:WTF does that mean?...zap!
Reply:Originally Posted by bassboy1On the periodic table, Silver's symbol is Ag, coming from the Latin word Argentum.  Argon, on the other hand, is Ar.
Reply:I have qualified and used He/Ar mixes for carbon steel. I have qualified used Ar/H for stainless steel. I have even used Ar/H on 1/4 and thinner carbon steel.
Reply:I haven't, but my buddy loves it on small beads and thing mild and chrome moly. I can see the difference in his weld with and with out. Says it pinpoints the arc better.
Reply:It does get the weld hotter, regardless of polarity, or process. Mig w/ it might not work well for different reasons, but yes it will work. For some mig work, they tell us to use a trimix gas. Helium is awful expensive to throw at a mild steel weld, or a SS weld for that matter. I have had a mixer set up and welded some other parts; just switched from AC to DC and went after it. No problem, it welded fine. It's just not practical for the cost.And then, after so much work...... you have it in your hand, and you look over to your side...... and the runner has run off. Leaving you holding the prize, wondering when the runner will return.
Reply:Thanks to u allG
Reply:Originally Posted by weldingliferI have qualified and used He/Ar mixes for carbon steel. I have qualified used Ar/H for stainless steel. I have even used Ar/H on 1/4 and thinner carbon steel.
Reply:Wow Pulser; that was great! One of those things..." I know it works, but I don't know why...." Now I know!And then, after so much work...... you have it in your hand, and you look over to your side...... and the runner has run off. Leaving you holding the prize, wondering when the runner will return.
Reply:Originally Posted by bassboy1On the periodic table, Silver's symbol is Ag, coming from the Latin word Argentum.  Argon, on the other hand, is Ar.
Reply:Originally Posted by pulserHydrogen is a diatomic molecule, but in the arc it is split into atomic H, absorbing a large amount of energy which is then transfered to the weld pool as the H2 molecules are reformed.  So, H2 helps in arc energy transfer, making a "hotter arc".  Apparently, H2 also acts to constrict the arc, but I don't know of actual measurements, or how this would happen.Helium greatly increases the heat transfer efficency in the arc because it's thermal conductivity is roughly 9X that of argon.  Additionally, since the inionization potential of He is higher than argon, for a given current and arc length the arc voltage of He will be several volts higher than argon, and since power, watts = amps x volts, there is more power to transfer with helium.
Reply:Metarinka,I am not aware how voltage affects the GTA plasma jet, as shown in the attached sketch, I understand it was a factor of the current flux and magnetic field in the arc.I have not seen, in my own work, or in the literature, that He produces a "narrower bead and deep penetration "finger"", as compared to Ar. I recall literature showing more melt volume for He, which makes sense since heat transfer efficiency is higher due to higher thermal conductivity, but the melt cross section tends to be similar, just larger, than Ar for a given set of conditions.As I understand it, energy from the GTA arc is primarily transfered to the workpiece by the impact of electrons, hence it is mainly a function of current as shown in the attached equation.  And thermal conduction of the plasma is the secondary factor. Attached ImagesPlasma Jet - 1.pdf (61.5 KB, 16 views)GTA Power Transfer.pdf (26.1 KB, 14 views)
Reply:Originally Posted by pulserMetarinka,I am not aware how voltage affects the GTA plasma jet, as shown in the attached sketch, I understand it was a factor of the current flux and magnetic field in the arc.I have not seen, in my own work, or in the literature, that He produces a "narrower bead and deep penetration "finger"", as compared to Ar. I recall literature showing more melt volume for He, which makes sense since heat transfer efficiency is higher due to higher thermal conductivity, but the melt cross section tends to be similar, just larger, than Ar for a given set of conditions.As I understand it, energy from the GTA arc is primarily transfered to the workpiece by the impact of electrons, hence it is mainly a function of current as shown in the attached equation.  And thermal conduction of the plasma is the secondary factor.
Reply:Sorry dude,it means that Metarinka and I agree that helium makes a "hotter weld", but we don't agree how it does it.
Reply:Pulser,wow that's a handy sketch thank you!now as towards your question on the penetration profile. The higher voltage of helium at the same arc gap will create a larger magnetic force and directly correspond to the plasma pressure. I will be the first to admit that I'm no plasma and arc physicist so I could be wrong. The plasma pressure of an arc is variable but  I don't have a strong understanding of all the factors that play into that voltage is a factor. now the deeper penetration profile I can personally attest to, as we specifically use it in fixed torch semi auto and auto GTAW welds to get deeper penetration. I'll have to look through my books to find a picture. It looks more like a nipple than a finger but helium produces a narrow bead at the center of the arc as opposed to the more round eliptical shape of Argon.  The deeper penetration as I understand it is a function of the current density, convective currents and plasma pressure present during welding (convective currents aren't directly affected by shielding gas). Now about the heat input  As quoted:"Then we should transform the expression of Henry into the expression derived by SCHOECK et al in 1963 [1], saying that the total energy input Q is:Q = I (Va+VKT+Vw)I  = Welding currentVa   = Voltage dropVw   = Anode work functionVKT  = Thermal plasma energy...This expression has now been used by GLICKSTEIN [2] who has thoroughly investigated the physical background of different thermal effects of different shielding gases (Ar, He, Ar+He, He+Ar, Ar+Al-vapour, He+Al-vapour).Hence, let me hereafter state what GLICKSTEIN has found out by his own and by using the results of many other famous researchers.The secret lies in the gaseous properties of Argon and Helium. It is well-known that the main portion of the kinetic energy in arc welding is carried by electrons. Thus, according to GLICKSTEIN, it may be supposed that the current density distribution ‘determines the distribution of the energy input to the weldment’. He explains that Argon has – due to the lower ionisation potential vs. Helium – a ‘higher electrical conductivity at lower temperatures’. Anyway, Helium again has a lower mass and thus, he describes Helium to have a higher thermal conductivity vs. Argon. GLICKSTEIN has investigated a 100 A GTAW arc and has calculated – based on experimental measurements – the electric fields for both gases as 15 V/cm for Helium and 7.7 V/cm for Argon. Thereby he explains the ‘broader and higher arc discharge temperature distribution’ for a Helium arc compared with an Argon arc. This, due to the so-called ‘energy source term’ is based on the temperature dependent thermal conductivity (sigma) and the square of the electric field (E) or:Sigma(T)*E²what again is explained to increase the radial energy transfer. And this, we all know from the practice, is true, since a Helium arc shows a more ‘diffuse’ appearance compared with an Argon arc. Thus, we may confirm what such great people like GLICKSTEIN et al have theoretically established long ago.So far so good. To explain the differences in both energy and temperature distribution in a 100 A Argon and Helium arc it is provided that the values for the variables in the first expression, as stated above, are:Va_Ar ~ 2.99 eV (anode drop for Argon)Va_He~ 4.71 eV (anode drop for Helium)Vw   ~ 4.5 eV (work function of Tungsten)VKT  ~ 1.2 eV (thermal plasma energy)Now one can calculate:Q_Ar   = I * (2.99 + 1.2 + 4.5) = 8.69 * IQ_He   = I * (4.71 + 1.2 + 4.5) = 10.41 * IFor a given current one obtains thus an approximately 20% higher thermal energy input for a Helium arc compared with an Argon arc.So, one may conclude, that no really significant difference in the ‘temperature’ between an Argon and a Helium GTAW arc has been found. But a quite significant distinction in the energy content of both arcs could be evaluated. To emphasize this, I would like to directly quote GLICKSTEIN [2]:“The greater energy input associated with the helium welding arc compared to an argon arc at the same current does not result from a ‘hotter’ temperature as commonly believed. Experimental measurements of the electric fields and analysis of the potential drop associated with each gas discharge show that the greater energy of helium can be attributed to a larger anode drop potential relative to that of argon arcs.”Nicely said and truly hitting the famous nail on the head.Isn't it good that the world has so tremendously intelligent people like Glickstein, Schoeck et al..?StephanReferences:[1] Schoeck, P.A. (1963), “An investigation of the anode energy balance of high intensity arcs in argon”, in: Ibele, W. (Ed.), Modern developments in heat transfer, Academic Press, New York and London, p.353-478[2] Glickstein, S.S., (1980), “Arc modelling for welding analysis”, in Arc Physics and Weld Pool Behaviour Conference Proceedings, Vol. 1, 8–10 May 1980, London, The Welding Institute, Cambridge, p. 1-8  "while it reaffirms what we both know (helium is hotter) the important difference is that thermal energy for plasma is considered relatively constant (or negligible) between the two gasses. The coeff of thermal conductivity is 9x higher but the density is ~10x lower at 1 atm. which means that while it is better at transferring heat; the davailable molarity of helium to transfer it is much lower so they more or less balance each other out.  From the above numbers the percentage of heat input via the plasma is only ~15-25%  of total Q. Gases have too low of specific heats and thermal conductivity to carry a large amount of heat energy for the low densities found at atmospheric pressure.here's the post I stole the quote from:http://www.aws.org/cgi-bin/mwf/topic_show.pl?tid=23466
Reply:thanks puls'r for the well needed clarification.both of u guys are on another frequency to me, but i got my answer so go aheadG