Heat input matters bigtime for D1.5 qualification.  And they don't address heat input calculation for multiple electrodes that I can recall.

Hg

Jeff,
Sirs,

first of all, thanks for mentioning my little spirit as to perhaps being able to having an idea on that!

This is a great question, and when I have read it first, I have thought by myself: "Wow, DC + AC combined! Extraordinary interesting!" Due to lack of time (lots of work currently) I have "stored" it on my "brain's hard disc" to surely coming back when I would have the next best opportunity. Now I am going to have it...

I guess you know me by now. I have deeply thought about, and to write some words on that was one of the truly hardest decisions I had to make since I have the honor to be a proud member of the AWS-Forums. As I have mentioned once. "Heat Input" is one of my "TOP-favorites" and perhaps this was the reason for so strongly hesitating with a response. Why..? Quickly said. Due to the outstanding complexity of each question treating "Heat Input" in arc welding. As far as I personally can think back. Each single question treating the simple term "Heat Input" had to be handled with lots of lots of mathematical formulae to be considered to find an adequate response on that specific question. And thus I hope imploringly to be able to humble try to response on "rtrautmans" post. I'll humble try...

However, those of you, who may have the will to read what I write, may forgive me when I start with some very fundamentally, this means, how the term "ENERGY" with particular respect to arc welding can, or should, be seen. Jeff is - once again - absolutely right when he mentioned: Quote "(...)what the difference is in terms of real energy and metallurgical reactions(...)" Unquote. When I speak with colleagues discussing particular issues of "Heat Input" I can recognize, that there are often "some" differences in the generally understanding of terms, used to describe the specific regimes of how an arc affects the material. This "fundamentally" treatment has somewhat to do with the way of how I have started to approach myself to "rtrautmans" post and thus I hope you may understand what has come out finally. Thanks in advance fellows! 
 
I am honest, I am far away from being any kind of Submerged Arc Welding expert but much more a layman in both theory and practice. Nonetheless, I have heard and remind to have learned anytime, that a combination as the mentioned one is not that seldom in using this welding process, in particular, please correct me when I'm wrong, when using the leading- and the trailing wire in rather less distances to each other and thus to reduce magnetic arc-blow phenomena.

And I am sure, although I admire the colleagues using this great and long-term approved welding process SAW, I will likely never be a member to this group of welders, since I guess that I likely never will learn it in the future.

What however interests me even more is the interesting theoretical coherence between DC and AC Current in SAW Tandem Welding and if this combination might lead to any kind of difference in the theoretical "Heat Input" (what ever this may be).

My second thought was then to try to remember what I have learned "somewhat ago" when talking of, or calculating the performance of an alternating current. And I remembered to have learned that the Power (P) equals the Voltage (U) multiplied by the Current (I) multiplied by the angle "alpha" divided by 360°, or:

P[W] = U[V] x I[A] x alpha / 360°.

That was when I have started thinking about what "alpha" stood for and which kind of role it played when they taught us about "Alternating Current" that time. I reminded that "alpha / 360°" had something to do with a value of momentum efficiency, since it shows a specific share (angle "alpha") of a circle (360°) which again is the basis of sinusoidal curvature which represents the sequence of an ideal alternating current or voltage, respectively.

Please see also the attached «Alpha.jpg».

What I have recalled as well was the fact, that the effective power of an Alternating Current is being calculated as the product of the Voltage by the Current by something called "cosine phi". This, as far as I remembered, stands for a special relationship between the primary side windings (supply voltage side) of the transformer and its secondary side windings (welding side). By more or less great stray losses between these both sides, transformers have a more or less large loss of power. This means, that the power on the secondary (welding) side of the transformer is even lower than that on the primary side, or, the tapping current of the secondary winding of the transformer is lower that of the primary winding. The fraction of what can be received from the primary winding of the transformer (normally < 100% or < 1) is called "Cosine Phi".

And then... my memory left me!

What I have recognized personally as well was, that as often I have very interested tried to follow the many progressing theories and calculations on "Heat Input", astonishingly I have never known a case where AC was required to be calculated.

Hmmm, quite funny...

O.k. due to my admitted lack of electrotechnical knowledge on this necessary detail, I have searched for some information on how AC as welding current should be possible to calculate. What I have found out was a lot far above my head but... that although there are very peculiar details to be considered when calculating very specific (electrotechnical) values of the AC regime, there is a good news for us welders. Mainly because the "secondary welding circuit side" AC is not a 3-phase AC (I avoid to describe the formula for calculating the power of it) but a single phase AC and thus the relations are quite simplified as to be seen subsequently.

When basically assuming an arc to having an electrical behavior of being purely OHMIC (no inductive a/o capacitive shares) we do have a great benefit. Why..? By having a look upon the (pure sinusoidal single phase) AC specifics, the  AC-power in an arc can thus be calculated by using the same formula as being used for DC-power, since the mean value power of an AC arc is ½ of its maximum power and this again equals the power of a DC arc, or:

P[W] = U [V] x I[A]

Please see also the attached «Sine.jpg».

This is pretty good - I thought by myself - since this enables us to use the formula as been already proposed by "rtrautman" and subsequently discussed by you all. But firstly...

"Efficiency" = "Arc Efficiency"?

Before resuming we should consider an important fact. Each burning arc is - of course - accompanied by energy or heat losses. This means that the total (electrical) power input P = U x I is not completely transferred into the workpiece but is being reduced by an almost endless row of different and sometimes quite intricate physical factors. But easily and in general one can say, that there is only a fraction of the total power used for creating an arc, melting the filler, melting the base material, etc.

What does this mean..?

Firstly one has basically to consider that in every (electrical) arc welding process the electrical power, tapped at the secondary winding side of a transformer (being then eventually rectified,...), can be simply calculated by using the mentioned formula above, which is P = U x I in [W].

This is the total power.

The arc efficiency is - as mentioned above - now defined as that fraction of the total process energy, being delivered to the workpiece a/o weld deposit.

In other words:

·  Heat Input to Workpiece / Total Power Input, or...
·  P (Power) = U (Voltage) x I (Current) x eta (arc efficiency)

Due to the arc efficiency is - as already mentioned - (normally) only a fraction of the total power input, the result of the formula above must be lower than 100% of the total power input, or even < 1.

However, when using the formula stated above, we have calculated the (momentum) power, considered to be reduced by energy losses and (mainly) being transferred into the workpiece. Normally though, the values of voltage and current are not constant under real welding conditions, which means that for finding the total "mean value" power we would have to calculate in a different way. But this - although important - should not be treated further herein. This means everything else we are talking about founds on the basic solution "Power = Voltage x Current" or "P = U x I" and I ask your understanding for this simplification.

Nonetheless I would like to treat very briefly another and not negligible fact. Therefore I would like to ask the following question:

"What do we really know, when we know the electrical power P = U x I?"

What does this expression provides us really with respect to a welding arc? I venture to say: "Not as much as we should know for continuing our discussion!" Why..? Well, what we know is a value for the electrical power, yes. And by using this value and multiplying it by the fraction of the value being used for "Welding" (arc efficiency) we do also know what amount of power is being used for "affecting" filler- and base material, yes.

And..? What now..? I do really hope you may agree with me, when I say: "This is not that much." Since as far as we do not know what this value means for the conditions of the workpiece subsequently to welding (mechanical properties,...) we do actually know "nothing". You surely know already what I mean! Yes. Since we have no idea of the "work" the arc has accomplished when carrying out its tricky interactions with the workpiece, we do have actually no idea as well on this work affects the base materials properties. What is being missed..? Of course! The time "t"!

As long as we have no result of where the time is playing a role by multiplying our voltage by our current within our "energy-calculations" we have likewise no idea of how the arc energy does affect our material. In other words, when we have a "theoretical" electrical power value (P = U x I) and we do not use it over a specific time (t) by igniting an arc which delivers the fraction (P = U x I x eta) towards the work, we can be sure to have no kind of any influence on metallurgical effects, etc. within our base material.

Using now the time as - as we have heard - crucial factor for calculating the "energy (E)" what nothing else than another form of "heat" (Q) we are now receiving the equation:

Pt = E (Ws = Joule) = U (Voltage) x I (Current) x t (time)

or (for our arc)

Pt = E = U x I x eta x t

to thus complete the things by now knowing the arc power, acting to affect the materials metallurgy, microstructure, etc. By the way, when we speak of calculating the "Heat Input" we can find our "energy" again as well by having a look upon the unit "Joule" [J] which is nothing else than "Watt-second" [Ws].

Everything else - pretty important but however - I would like to avoid to treat herein, to not complicating the entire subject unnecessarily.

But...

It is well-known that different arc welding processes exhibit different arc efficiency values, quite depending to their thermal energy losses due to radiation, convection, etc. Furthermore one should consider, that different Researchers have found and measured different arc efficiency values from the past up to now! This - I hope you fellows would agree with me - doesn't even simplify the entire subject.

Hmmm, but what does this actually mean? Different indications for one and the same welding process..? Different Researchers = different efficiencies,..? Who is even blessed to keep here the complete overview and how can I be sure that the "Heat Input" been calculated by me - Stephan - is quite the same as if it has been calculated by e.g. "rtrautman" or any other forum member?

As mentioned, as just for nearly any arc welding process as well for the Submerged Arc Welding there are different values known in the literature to define this particular and important fraction of energy. These values lay between 0.75 (RYKALIN), 0.9 (http://www.sandia.gov/soar/Pdf_docs/Ener_trans_effic.pdf) up to 1 (SCHULZE-KRAFKA-NEUMANN).

In other words this does imply that between 75%... ~ 100% of the total power input are used for the process as to be converted into effective arc energy, or, approx. 100% of the thermal energy are used for an interaction with the base material for melting and depositing. This is mainly being founded on the specific character of one (or even more) SAW-electrode(s) burning in a complete covered flux cavern. Hereby again the losses by radiation, convection, weld spatter,..., are extremely low. And exactly this fact we should use as an important advantage for our further discussion.

Would this mean in fact that it might happen that depending to the arc efficiencies being used for calculating the "Heat Input" the results might  differentiate? So what might be the correct values for calculating the "true" Heat Input? Let us make an example and let us calculate two values of arc power by using different values of arc efficiency.

1.  Example:
 
·  U = 25 Volt
·  I = 200 Ampere
·  Eta = 0.75

P   = U x I x eta
  = 25 [V] x 200 [A] x .75
  = 3750 [W] = 3.75 [kW]

2.  Example:

·  U = 25 Volt
·  I = 200 Ampere
·  eta = 1

P   = U x I x eta
  = 25 [V] x 200 [A] x 1
  = 5000 [W] = 5.00 [kW]

Of course one can easily see, that there is a significant difference in the performance - even 25% - of both "virtual" arcs. When we would now have to subsequently calculate the "Heat Input" of the arc we would of course get two different values of "Heat Input", only by using two different values of arc efficiency (eta).

Now we are at the point where I would like to quote our great colleague Jeff as he stated:

(Quote): "You must be sure to bring apples to apples(...)" to "...compare apples with apples(...)" (Unquote).

This means that one should use the values for efficiencies just as being fixed by the national or international standardization organizations or to follow the recommendations of the national or international welding organizations. Only this can ensure that "apples will be brought to apples" and the obtained results would have the ability to be compared to each other.

Please allow me to subsequently simplify the whole subject we are talking about. Let us assume that the arc-efficiency of the SAW-process could be fixed at "1", what implies that the arc is so completely covered by the flux, that all minimal heat losses could be neglected and thus we are working as if the complete electrical power would be converted into the arc-efficiency and transferred into the workpiece. Furthermore let us assume, that we are using a sinusoidal wave formed Alternating Current, which allows us to use the same formula as for Direct Current (P = U x I).

This would mean that our Alternating Current Arc Energy could - as mentioned, just theoretically and just for simplifying - be calculated by:

E = U x I x 1 (eta) x t or:

E = U x I x t

and thus we could normally calculate the "Heat Input" just exactly as "rtrautman" and Jeff have described, namely by:

Heat Input Q = U [V] x I [A] / v [cm/s] or even converted:

Q = Volts x Amperes x seconds / cm  = [J/cm]

Now I should normally finish my reply. Hmmm, actually yes. Since it could be seen that - under real conditions - it might be feasible to calculate the energy of a single phase sinusoidal AC arc by using the identical formulae as being used for a DC arc, everything should be said and done. Or rather not..?

Well, from my humble point of view not at all. Since there is one item worth to be considered a bit more in detail. And this item is being mentioned by Jeff, when he says (Quote): "And you calculate them separate if the leader puddle solidifies. You calculate them combined if the trailer hits the leader puddle while still molten." (Unquote). And as well when HgTX says (Quote): "And they don't address heat input calculation for multiple electrodes that I can recall." (Unquote). This is - from my personal standpoint - a great deal! I must request your understanding for not knowing what D 1.5 is. But what I quite accurately know is that it is an intricate undertaking to calculate the "Heat Input" for multiple electrode arc welding processes. Not only by considering the very different conditions in Joint-Geometries, sheet metal- or plate-thickness, welding position,..., but also by many items to be considered with regard to welding speed, physical base- and filler metal properties, etc. Thus every calculation should - at least from my standpoint - be approved by practical trials and measurements. This should be a must, since the many different variables affecting the final welding result and much more the final joint- and base metal properties should be considered by using the practice for "calibrating" the theory. As far as I know the steel manufacturers or mills - at least the greater ones - are the ones who have investigated all material's properties with respect to its workability. Amongst others likewise the "weldability" of the material and thus have established the recommendations for welding the material in a proper way (CCT-Diagrams, t8/5 data, etc.

But however, what about the welding with multiple arcs or electrodes, respectively? What about when no data is available for such kind of welding process? O.k. what Jeff has recommended I personally would like to 100% agree with. Calculate the "Heat Input" of both arcs separately if the leader puddle solidifies and combined when the trailer puddle hits the leader puddle.

But what I asked myself was: "When does the trailer- hit the leader-puddle? When is the leader puddle solidified? Isn't it depending again to again different variables? You see, questions over questions. Perhaps unnecessary to reply since being negligible, but - at least fro my personal standpoint - interesting to be considered.

To not overwork the entire subject, please allow me to just having an additional look upon the "puddles".

To know if the trailer- hits the leader-puddle one should consider:

1.  The length of the leader-puddle
2.  The spatial distance between both wire-electrodes or arcs, respectively.

To adequately response the first, one should consider again by which factors the length of the weld pool is being influenced. And now it would get quite intricate, since there is a tremendous number of variables again which could affect the length of the weld puddle. All these variables will likely never be comprehended completely due to the number of variations in material (base and filler) combinations, geometrical combinations, etc. is nearly infinite. But however there were investigations been accomplished in the past to find a theoretical basis for a feasible calculation of the weld pool length. Due to the SAW process has the highest arc efficiency of all arc welding processes and additionally the electrical parameters (current + voltage) are chosen to be "high" one could assume that the SAW weld pool has the comparative greatest lengths. To now calculate the weld pool's length without a senseless search for variables anyhow never to be known completely, one had to neglect many of the unknown theoretical properties in coherence with arc and base material. Furthermore it should be reasonable to find a way of calculation being easy to handle by the welding practitioner. These requirements were met, by establishing a relatively simple expression:

·  Length of the weld pool (L) = specific coefficient (p2) x U (Voltage) x I (Current) or:
·  L = p2 x U x I

First we can see, that our "normal" power (U x I) can be found again as well in this expression. Good for us. But what is "p2"? Well, p2 is a specific coefficient (factor of proportionality) having the unit length/power (mm/kW) and which has to be evaluated by practical experiments to make sure that the theoretical material parameters are considered in a wide extent, at least practically. Practical investigations been carried out by LJUBAWSKI and LASARJEW have yield the subsequent results for the SAW process using unalloyed wire electrodes on unalloyed base materials:

·  Current (I) < 1200 [A] -> p2 = 2.3... 3.6 mm/kW
·  Current (I) 1200... 3000 [A] -> p2 = 2.3... 2.8 mm/kW

An important role herein plays however the welding speed, which has to be considered appropriately. This is what makes the practical surveys that important and making the values above only being a "rule of thumb" to be approved by the practice. Nonetheless by now knowing approximately either the theoretical or practical length of the leading weld pool one can approach if the trailer puddle may hit the leader puddle and thus to calculate the "Heat Input" commonly or separately. As I said, I am truly no SAW welder or even expert. Thus I do not know the "normal spatial distances between the wire electrodes when multiple SAW process is used. I guess however, that this distance may be larger than the leader puddle's length and thus a separate calculation could occur. But I do not know. If however the spatial distance between the wire electrodes exceeds the leader weld pool's length I do not know how this fact might influence the specifics with regard to a particular consideration of the interpass temperatures. This means, when the distance between both wires is high enough and the leader puddle would have solidified "long" before the second wire passes the already solidified leader weld pass, how far should this be considered as a "preheating" only by the fact of an increase of the interpass temperature? I know that there were some interesting investigations been conducted by RADAJ based again on results found out by RYKALIN who has differentiated between "Multilayer" and "Tandem-Welding". Here - as far as I remember correctly - were distances of < 0.1 m between the arcs considered as to having a "preheating" function and distances between the arcs of > 1.0 m were considered as to calculate the "Heat Input" separately. But once again, only under restriction, since I do not know if I recall really correctly. I would have to investigate this again...

O.k. coming slowly to the end...

To make what I wrote "complete" although I know what we are talking about will likely never be completed.

Once I was asked by a good colleague of mine if I could have a view on GMA Tandem-Welding application and, if possible, to calculate the "Heat Input" for this particular application. That was the point in time when I have first busied myself with this tremendous interesting subject. At that time I have quasi followed Jeff's recommendation of calculating both arc performances - and just these - by adding them together, which was founded on the assumption that both arcs act within a common weld pool. This again I assumed due to the fact, that both wire electrodes had a spatial distance of approx. 14 mm. Basically I had to make my approaches on the application at that time by knowing that two "conventional" (spray) arcs had been used. Due to a common weld pool was assumed to be created by both single arcs, one had to calculate the values of both the "Leading" (Qa l) and "Trailing" (Qa t) arc first and subsequently add them to obtain the total process energy (Q a) by:

Q a = Qa leading + Qa trailing

This can be relatively easy executed when a DC is being used for both electrodes, since then one can work with the formulae mentioned above. Namely

Qa = (U x I x eta [leading]) + (U x I x eta [trailing])

Mentioned by the way, when a pulsed arc is used on both leading and trailing wire electrode, one has to consider the altering values of both voltage and current (both as functions of time) over the time the arc burns. This requires a different method of calculation. For those ones who are interested, please see also the attached «Q a.jpg».

As mentioned, at that time I had to try to find a differentiation of the efficiencies between the "Heat Input" of a conventional GMA Tandem Welding process (using two 1.0 mm diameter wires) compared with a conventional Single-Wire GMA-Welding process (using a 1.6 mm diameter wire). It was important here to find out if the GMA Tandem Welding process should have a lower "Heat Input" compared to a Single Wire GMAW process although the addition of both current and voltage in Tandem Welding would exceed the value of both in Single Wire GMAW process. An important fact herein plays, as already mentioned, the welding speed which is considered to be the denominator in our "Common Heat Input" expression. In GMA Tandem Welding relatively high welding speeds can be used compared with Single Wire GMA Welding.

For the GMA Tandem Welding I have used the following values:

·  U1   = 22 Volt
·  I1   = 250 Ampere
·  U2   = 20 Volt
·  I2   = 220 Ampere
·  Dia.  = 1.0 mm (both)
·  eta   = 0.85
·  v   = 40 mm/s

Then I have calculated the "Arc Energy" by:

E   = (U1 x I1) + (U2 x I2) x eta
  = (22 x 250) + (20 x 220) x 0.85 
  = 8415 [W] = 8.415 [kW]

Subsequently I have calculated the "Heat Input" for the Tandem Welding process by:

Qa [J/mm]  = E x t / v
    = 8415 [W] x 60 [s] / 2400 [mm]
    = ~ 210 [J/mm] = ~ 0.21 [kJ/mm]

Then I have used the following parameters for the Single Wire GMA Welding process:

·  U     = 41 Volt
·  I     = 400 Ampere
·  Dia.   = 1.6 mm
·  eta   = 0.85
·  v    = 10 mm/s

"Arc Energy" was calculated by:

E   = U x I x eta
  = 41 [V] x 400 [A] x 0.85
  = 13940 [W] = 13.94 [kW]

"Heat Input" was calculated by:

Qa [J/mm]  = E x t / v
    = 13940 [W] x 60 [s] / 600 [mm]
    = 1394 [J/mm] = 1.394 [kJ/mm]

I guess, as far as I have calculated correctly (so please correct me if you may disagree) hereby one can see remarkably that although the "pure" Tandem welding parameters (voltage + current) do exceed the Single Wire Welding parameters, the Tandem Welding process has however a lower "Heat Input" value compared with the Single Wire process. Only by being able to use the Tandem process under higher welding speeds, which decreases the "Heat Input" as the ratio of Energy to unit length. I mean this is a quite interesting example for how one can reduce the "Heat Input" by using "High Performance" Welding processes being simultaneously "High-Speed" Welding processes.

Last but not least. Meanwhile we have conducted internally calorimetric measures by using Tandem GMA-Welding compared with Single Wire GMA Welding. These trials - which I would not like to treat the extensive details of herein - could nonetheless approve that when using wires comparable to ER 70S6 and welding bead on plate welds onto unalloyed steel base material (S 235 according to EN 10025) having dimensions of 250 mm x 50 mm x 5 mm, the Tandem Welding is approx. 1.9 as efficient as the Single Wire process. This again leads me to the assumption, that the way of calculating the "Heat Input" of a Tandem Welding Process where both arcs acting on a common weld pool are - at least specifically - correct, by adding both arc energies and divide the sum by the welding speed.

O.k. that far my humble contribution on "rtrautman's" interesting post. As always I request your kindly correction when I've been on the wrong path...

EDIT:

There is a detail which I have thought about and which doesn't let me settle down.

I am no electrical engineer thus please correct me when I am wrong. I considered the following subject. I guess in SAW we are using a 3-Phase Alternating Voltage being transformed to generate a relatively low Alternating Voltage but a high Alternating Current.

On the other hand we know, that the 3-Phase Alternating Voltage or Current respectively, have to be calculated by mandatory integrating the phase shifting by using the square root of 3. Thus we would receive the following expression:

P = U x I x eta x square root 3

This again we have heard is - with respect to the primary winding's side of the transformer - not the effective power to be used for the arc efficiency on the secondary side of the welding circuit, but only the power without considering "Cosine Phi". Now I have thought about the following detail. When "rtrautman" is using AC for his trailing arc and the efficiency of this arc is "normally" to be seen as 1.0 as already described above, then the total efficiency of the AC-arc however must be less than "1.0". Why..? Well, when we know that the electrical efficiency on the secondary side of the welding circuit is directly depending to the relations on the primary side (winding) of the transformer, and we also know that the secondary side is lowered in an amount being expressed by "Cosine Phi", we should have to calculate the true efficiency of an AC-Welding-Arc - which is of course generated on the secondary side of the welding circuit - by:

P = U x I x eta x square root 3 x cos phi

or

P = U x I x eta x 1.73 x cos phi

Cosine Phi again - which is the factor standing for the reduced secondary efficiency of an Alternating Current - should be recognizable on the power rating plate. At least in Germany one can recognize it there.

Finally this would mean to me that the AC-arc power should be lower than the "normal" DC-arc-power namely exactly by that difference between "1.0" and "Cosine Phi" (e.g. cos phi 0.8). In this case this means, reduced by 20%.

Hope that I am right by thinking so, but however, finally I am 100% with Jeff who said "bring apples to apples" to "compare apples with apples".

Thanks friends..!

My very best regards to you all!
Stephan