A common question that is asked when manufacturers are trying to determine costs associated with welding is: How many pounds of wire can I deposit per hour? Or, how many pounds of wire can I deposit at a specific amperage or wire feed speed? There are charts to help you determine this, but it’s always good to have a simple formula.
Before we get into this calculation is very important to make a distinction between deposition rate and melt-off rate. A lot of people in our industry use these two terms interchangeably and that is not correct. Melt-off rate is how much wire is being consumed and does not take into account the efficiency of the electrode. Deposition rate is how much wire is actually converted into weld metal. Solid wire efficiencies can range from 88 – 98%, so you can have the exact same melt-off rate but considerably different deposition rates if you are welding at the same wire feed speed but with a different mode of metal transfer.
So now to the formula:
Deposition Rate = 13.1 (D²) (WFS)(EE)
D = electrode diameter
WFS = wire feed speed (in/min)
EE = electrode efficiency
13.1 is a constant used for steel and it is based on steels density. This same constant can be used for stainless steel wires as it is only 1/1000 difference. Aluminum on the other hand would have to be a constant of about a third that of steel, or 4.32.
Electrode efficiencies for solid wires can vary depending on the mode of metal transfer. We are not going to get into the specifics of how to achieve each at this time. Below are typical efficiencies for each of the modes.
Short-circuit Transfer: 90-93%
Surface Tension Transfer: 98% (STT is a trademark and of the Lincoln Electric Company)
Globular Transfer: 88 – 90%
Axial Spray Transfer: 98%
Pulsed Spray Transfer: 98% is typical, but can be lower depending on the parameters and power source
Example: A customer wants to know the deposition rate of an .052 ER70S-6 wire. They are running 90/10 shielding gas at 320 in/min and 30 volts. We are in spray transfer mode at these settings.
Deposition Rate = 13.1(0.052)² (320) (0.98) = 11.1 pounds per hour
NOTE: This formula only works for with imperial units, if you are using the metric system simply follow the following steps:
- Calculate the area of the circle -> Radius squared * pi
- Multiply times the density of the metal being used
- Multiply times the wire feed speed
- Multiply times 60 (to get deposition rate per hour if wire feed speed was expressed per minute)
- Adjust any units to match (i.e if you used cm for the density or other units)
- Multiply by the efficiency of the electrode
How did you translate the deposition rate to pounds per hour when your wire feed speed is in inches per minute?
Hello Kelby. The formula provided in the above article does the calculation for you. But, if you want to do it the long way all you need to know if the diameter and the wire feed speed. If you take the diameter of the wire you can calculate the cross sectional area (A=πr2). If you know the speed in inches per minute you know the the length of wire by unit of time. So you are able to determine the length of wire fed and from here the volume (Area * Length). This give you a a number in cubic inches per minute being fed. You then take the density of steel (0.283 lbs/cubic inch). When you multiply those two units you get the result in pounds/minute which can then be translated to pounds per hour. Hope this helps.
Would my travel speed be the same if I am trying to use a weave with a dwell in the horizontal position?
If you have dwell time your travel speed would decrease; however, your bead would be bigger than just doing a stringer.
Hi, I can use this formula in the metric system? I also want to know which literature is this formula. Congratulations on the site! I can get out of difficult situations with the knowledge that I have been getting.
Hello Matthew,
This formula cannot be used in the metric since the 13.1 constant is specific to the imperial system. The above formula comes from Lincoln Electric’s GMAW Guide as shown at the bottom of the article. This can also be found on the Procedure Handbook of Arc Welding (you can find a link to this on our Resources tab). The Procedure Handbook does include Metric System formulas. We will search for the metric formula and provide if found.
Thank you very much!
5.954*((Rod Dia mm/25.4)*(Rod Dia mm/25.4))*(mm/min wire feed/25.4)*electrode efficiency
This works with the Metric system – result is in Kg/Hour and inputs are in mm and mm/min
Thanks for sharing Rob!
I need to know how calculate the deposition rate has been placed in the vertical position during the SMAW welding
Please refer to the post on “Deposition Rates for Stick Electrodes” by following the link below
https://weldinganswers.com/deposition-rates-for-stick-electrodes/
If you know the current you are using simply used the formulas provided for the type of electrode you are using.
Pls provide book for my reference.
Which book do you need? If you want the free Guide on Writing a Welding Procedure Specification simply enter your name and email address on the side bar and then confirm your free subscription when you receive the welcome email.
Does this formula apply to stainless steel GMAW wire as well?
Yes, you can use it for stainless. The density of stainless steel is almost identical to that of carbon steel. If you were running aluminum then this formula would no longer be applicable.
can i use this formula for GTAW?
The only way you could use it is if the filler metal was feed at a consistent speed. This is almost impossible manually. If you use a cold-wire tig feeder (or hot wire for that matter) with a consistent speed you can certainly use this formula.
Then how can we determine which type of metal transfer in TIG for electrode efficiency?
The modes of metal transfer only apply to wire processes. With GTAW you have an arc and are adding filler to puddle. The filler does not conduct electricity (it can in hot wire TIG) in most cases. The electrode efficiency is how much of the electrode is consumed and turned into weld. Since GTAW generates not spatter (at least it shouldn’t) you should have a 100% efficiency. In hand held applications you will end up with 1 to 2 inches of filler metal which is to small to handle and add as filler. If you have a 36-inch cut length your efficiency would be 94 to 97%.
If you are asking about deposition rate then that depends entirely on how fast the welder feeds the wire. On automated applications such as cold or hot wire tig applications you can have a consistent wire feed speed. If that is case you can use the formula supplied in this article. If it is handheld you’ll have to time the welder and then measure how much of the cut length was used. Then with simple math and using the density of the filler material you can arrive at an “estimated” deposition rate.
At what point in this formula does the Voltage and Amps come into play?
Neither voltage or amperage come into play. Amperage is a function of the wire feed speed. Wire feed speed should remain constant but amperage will vary based on contact tip to work distance. This is why we use wire feed speed and not amperage in the calculation.
Hello, I am using the SAW process with 2 welding heads my controllers only show amps and volts, could you show me or is there a conversion out there to take amps and give you wire feed speed? I’m running 765 amps on one and 565 on the other. thank you
Hello Garrett. There are charts that can get you close, but amperage and wire feed speed are not perfectly related. At the same amperage you can significantly different wire feed speeds depending on your contact tip to work distance.
Take a look at this SAW guide from Lincoln Electric. Look at page 15 of 54 for a chart that compares WFS to amperage.
https://www.lincolnelectric.com/assets/global/Products/Consumable_SubmergedArcConsumables-Lincolnweld-LincolnweldL-50/c550.pdf