Welding Math (and Physics)

For Welders, Welding Students, Welding Instructors and Others Involved in Managing Welding Operations.

Understanding the fundamentals of what makes a weld will assist in producing higher quality welds and when assessing welding problems.  If you are using or planning to use welding Robotics this information is essential.

MIG WELDING

What Causes Welding Wire to Melt? 

This is the question I ask when giving a technical talk on the subject.  Many answers are offered.  Here is what it is NOT;  the “hot arc”, radiation from the arc or the wire passing through the arc.  Two phenomena are primarily responsible for wire melting.  A simple test will explain one of them: 

1. BE VERY CAREFUL, ONLY CONDUCT THIS TEST WITH INSTRUCTOR SUPERVISION--AND STAND BACK THE WIRE MAY "EXPLODE!!"   (Use a Stick welder.  With the machine turned off; clamp one end of a 3 foot length of 0.035 welding wire in a stick electrode holder and the other under a work ground clamp .  With the stick welder set for 150 amps, turn it on (be careful to stand back). The wire will quickly turn red, then white and then it will melt like a fuse. Note the short time that took.  That demonstrates one of the causes of wire melting during welding.  As the wire passes from the end of the torch contact tip to the arc, it is carrying all the welding current and becomes very hot.  It starts at room temperature and can exceed 500 degrees F before the arc forms at the end (depending on the "sickout," the distance between the tip and the workpiece.)

2. The second reason it melts is that current leaving or entering a surface, be it wire or hot puddle, requires a given amount of energy for the electrons to enter or leave that surface.  This energy, generated at the surface, melts the already hot wire.  Therefore assuming Electrode Positive this is referred to as Anode Potential (also called Work Function and measures as voltage) and is equal to the Amps x Anode Voltage.

The following equation obtained from Reference (1) defines the relationship:

Wire Melting Rate = a x Amps + b x Wire Stickout x Amps ²

Where "a" and "b" are constants and "Wire Stickout" is the distance from the torch contact tip to the workpiece measured in inches.

The values for "a" and "b" for 0.035 inch diameter carbon steel wire are:

a = 0.0086;  b = 0.000078

These two energy sources cause the wire to melt.  The first term (a x Amps) is the anode voltage times current and the second term defines the energy input due to resistance heating.

IMPLICATION OF THE EQUATION

One of the major implications of this relationship is that an increase in the stickout (at a fixed wire feed speed) amperage will decrease.  That has a significant effect on another parameter, weld penetration.

(Note, stickout is also called tip-to-work distance. The term "Electrode Extension" or ESO is a more accurate measure of the distance between the end of the contact tip to the top of the arc.  However for these calculations the stickout or contact tip-to-work can be used and is easier to measure.)

Point of Interest

The voltage you measure on a MIG welder is a combination of the:

1. Small voltage drop due to resistive heating of the wire (I²R)

2. The anode and cathode potentials (that needed to get electrons out of the wire and into the puddle; could be 1/2 the total measured)

3. The voltage drop across the resistive arc

WELD PENETRATION

Weld penetration can be determined by a simple equation defined some years ago by C. E. Jackson in Reference (2)

Weld Penetration (distance into the base material when making a weld on plate measured in inches) =

K [Amps⁴ / (Weld Travel Speed; ipm x Volts ²)]^0.333

For 0.035 inch diameter solid carbon steel wire, the constant K = 0.0019

Using these equations we find the following when we change wire stickout for 0.035 inch solid wire.  Assuming a fixed wire feed speed that produces 200 amps at 3/8 inch wire stickout:

Note:

1. With a fixed wire feed speed the amperage decreased from 200 amps with a 3/8 inch stickout to a low of 152 amps when the stickout was increased to 7/8 inches.  The resistance heating of the wire (the 2nd Amp² term in the equation) is a very efficient heating process.  Therefore the current needed to finish melting the wire as it enters the arc, becomes less as the wire is hotter with longer stickout.

2. However there is a reduction in weld penetration when varying stickout in a normal range from 3/8 to 3/4 inches is 25%!  If extended to 7/8 inches penetration decreases 31%.

Therefore it is very important to keep the torch stickout constant.  Also the shorter the distance from tip to work for a fixed wire feed speed the greater the penetration since current also increases.  When welding in the short circuiting mode it is often desirable to use a long contact tip which protrudes from the shielding gas cup.  This helps assure adequate penetration is maintained by keeping current higher.  It also helps visibility so the welder can stay on the leading edge of the weld puddle.

Weld Penetration Definition

For the purposes of this exercise, weld penetration is a measure of how deep the weld penetrates in a bead-on-plate deposit. Have a different wire than the 0.035 inch solid wire used in this example?  No problem.  In fact not only changes in wire type and size but also shielding gas and torch angle will alter the actual value.  You can generate your own constant K for what you are doing by making a bead-on-plate deposit, cutting a cross section and etching it.

Caution About the Use of This Approach

The above equations are designed to work within a practical range for normal MIG welding.  There are much more elaborate equations that take into account the variable resistively of material with temperature, other arc effects etc.  However within a range of normal operation the above approach will work adequately.  As mentioned, you may need to develop your own coefficients.  Make a weld, measure the depth of penetration and work backwards!  That is what Clarence Jackson did with thousands of data points!

Reference (1):  AWS Welding Handbook, Volume 1,  9^th Addition; pp 79

Reference (2): “The Science of Arc Welding” by C. E. Jackson. 1960 Welding Journal 39(4) pp 129-s thru 230-s

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Gas Pressure/Volume Calculations

Understanding the relationship between gas pressure and volume will help when setting gas flow and understanding MIG shielding gas use and waste.

In 1662 Robert Boyle defined the basic relationship between gas Pressure and Volume.  He stated the relationship in mathematic terms as:

P1 / P2 = V1 / V2   (Where P, pressure, is measured as absolute pressure =

gauge pressure + 14.7 psi at sea level)

For example, in a MIG gas delivery hose if the initial absolute pressure is doubled the gas volume must be double.  That was until 140 years later when around 1800 both Charles and Dalton independently added temperature to the relationship!  They defined the following relationship:

V1 / V2  = T1 / T2  (Where T, Temperature is measured as absolute Temperature in degrees Rankin = Degrees Fahrenheit gauge + 460)

Or at a constant pressure, an increase in Temperature of 20% measured in degrees Fahrenheit yields only 3% increase in volume.  For most welding purposes and over the range of production temperatures there is not enough change in Volume with gas Temperature to consider it significant.

The general Gas Laws can be written as:

(P1 X V1) / T1 = (P2 x V2 ) / T2

It can also be written as P x V = (nR) T

where the constants nR can defined based on the gas.  We'll let you find details about the nR by searching the Internet since for our purposes ratio comparisons are all we need.

MIG WELDING EXAMPLE

Assume:

1. The pressure needed in the gas delivery hose from gas source to welder/feeder to flow 30 CFH is 3 psi. 

2. A regulator/flowmeter is used that utilizes an 80 psi regulator.

Therefore when welding stops gas continues to flow through the needle valve flow control  until the pressure in the gas delivery hose equals 80 psi.

Then the volume of excess gas in the gas delivery hose compared to the volume when welding will be:

V stopped / V welding = P stopped / P welding =

(80 psi +15) / (3 psi + 15) = 5.3 times the physical hose volume

Hose Expansion Causes More Excess Volume: Tests of a standard 1/4 inch diameter gas delivery hose showed it had a 13% increased volume due to the hose expanding with 80 psi pressure.  Therefore 87% of the excess gas stored in the hose is due to the increased Pressure and 13% of the excess due to Hose Expansion.  SEE MORE DETAILS.

Was This Welding Math Information Useful?  What Other Math Would You Like To See?  

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