Wrong Tim, but thanx for your reply.
I found a formula in the "Lincoln Electric Procedure Handbook of Arc Welding."
Approx. Melt-off Rate = (Arc Voltage x Welding current) /1000 = lbs/hr.
One of our directors had the book stashed in his cupboard. It's not a very accurate calculation, but it's all I need.
Thanx for your guys' time anyway. Keep well.
It sounds like you already have a fairly good idea of what you were looking for. From reading through the posts, I'm not sure I do.
But anyway... I just like to caution people about using deposition rates and similar info. You may already know all this but here it is, if not for your benefit, then for the benefit of others.
Most values are in weight per hour and it is easy to forget that a person cannot weld continuously for an entire shift, unlike a machine. Even machines might not weld continuously.
I've seen a lot of people assume the deposition rate can be multiplied by hours per shift for a value representing productivity. Actual productivity is best judged by the amount of acceptable and finished welds made by the end of the shift.
Chet Guilford
Hi Duncan
I find the equation above rather odd, because the deposition rate for GMAW will remain the same irrespective of the welding voltage. Obviously the welding voltage will only be variable within a useable range, but theoretically you can vary the welding voltage by 30% and still have exactly the same deposition rate!
I also have the "Procedure Handbook of Arc Welding", but could not find your equation in it. There are however numerous graphs that equate welding current and electrode size to both wire feed speed and deposition rate. (In GMAW these are almost synonymous, because once you have selected the wire feed speed, you have automatically selected your deposition rate. The amp and volt are not an issue. Obviously with a CV power source, your amps are related to wire feed speed, so this becomes easy enough to relate to each other.)
Let me know on which page you found the equation mentioned below. It would interest me to have a closer look.
Regards
Niekie Jooste
Dear Duncan,
I completely disagree with you !
Here are some figures that I've looked up :
case (1) :
* stick out (extension) 12 mm
* wire feed speed 5.8 m/min
* current 350 amps
* deposition rate 4.7 kg/hr
case (2) :
* stick out (extension) 18 mm
* wire feed speed 5.8 m/min
* current 320 amps
* deposition rate 4.7 kg/hr
case (3) :
* stick out (extension) 25 mm
* wire feed speed 5.8 m/min
* current 280 amps
* deposition rate 4.7 kg/hr
CONCLUSION :
By increasing the stick out, you will decrease the welding current !
The wire feed speed is in every case the same , resulting in the same deposition rate. So, in my opinion you can not calculate the deposition rate based on the welding current without taking into account the effect of the electrode extension or the wire feed speed.
But I do believe the "formula" gives you an idea of deposition rates for certain wire types as a rule of thumb.
As you see, this is rather complex. And than I'm not talking about flux cored wire or metal cored wire ... Did you also know that by changing the polarity at the same welding current you can also increase the deposition rate (due to the physics inside the welding arc) ... :-)
Hi Tim
What you have illustrated above is simply that the deposition rate is directly related to the wire feed speed. The rest of the information tends to cloud the issue.
I agree that it starts becoming complicated when trying to relate the deposition rate to the welding current and voltage, but in essence it is a very simple issue. The amount of wire coming out the front of the torch will be deposited. The only two variables you therefore need are the wire feed speed and wire diameter.
If however Duncan REALLY, REALLY, REALLY wants to relate the deposition rate to the welding current and voltage, his best bet is to relate it directly to the current, because there is a very close correlation between wire feed speed and current. Obviously there will be certain variations introduced, as you indicated, but he probably just needs some approximation for tender purposes or something similar.
Regards
Niekie Jooste
Fabristruct Solutions
The info is sometimes supplied in a ratio of #'s of wire vs #'s of deposited weld metal. Ie. FCAW, SMAW, etc. - part of the weight ends up in fume,splatter or slag to get chipped off and isn't really part of the final deposited weld material. I was looking at a wooksheet in ESAB's catalog that was to figure weld metal cost and one of the formulas was to figure Deposition Efficiency (DE= Weight of weld metal divided by weight of electrode used) ie if it takes 4#'s of wire to deposit 3#'s of weld metal the DE is 75%.
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