Welding math

Arc Energy and Arc Efficiency

Not all the energy an arc produces ends up in the steel, and the fraction that does depends heavily on the process, which is why the same gross heat input means different things under a flux blanket and under an open tungsten arc.

The formula

HI_eff = f x (V x I x 60) / (1000 x v)

Effective heat input equals the arc efficiency factor for the process multiplied by the gross arc energy per inch.

What each symbol means

SymbolMeaningUnits
HI_effEnergy per inch actually absorbed by the workpiecekJ/in
fArc efficiency, the fraction of arc energy that enters the workdimensionless
VArc voltagevolts
IWelding currentamperes
vTravel speedin/min

Why it works

Gross arc energy counts everything the power source delivers to the arc. Some of that radiates away as light and infrared, some convects off in the shielding gas, some leaves in the spatter that never reaches the joint. What remains is the effective heat input, the part that raises the temperature of the plate. Multiplying gross energy by an efficiency factor is the standard way to account for the difference.

The factors are process characteristics, not machine settings. A gas tungsten arc is open and radiates freely, so a substantial share of its energy escapes; figures in the region of two thirds are typical. Consumable-electrode arcs do better because the molten droplets carry heat directly into the puddle. Submerged arc does best of all, because the flux blanket traps radiation that would otherwise escape and returns much of it to the joint.

This matters most when a procedure written for one process is reused for another. Thirty kilojoules per inch gross from submerged arc puts far more energy into the plate than thirty kilojoules gross from gas tungsten arc welding, so a limit transferred between the two without adjusting for efficiency will be wrong in one direction or the other, and both directions cause problems.

Worked examples

Spray transfer gas metal arc weld

Arc voltage
28 V
Welding current
280 A
Travel speed
14 in/min
Arc efficiency
0.85
  1. Gross arc energy: 28 x 280 x 60 = 470,400 joules per minute.
  2. Divide by 1,000 x 14 = 14,000 to get 33.6 kJ/in gross.
  3. Apply the efficiency factor: 33.6 x 0.85.

Result: 33.6 kJ/in gross, 28.56 kJ/in effective
About five kilojoules per inch never reaches the plate. Thermal predictions built on the gross figure would overestimate the peak temperature and the width of the heat affected zone.

Gas tungsten arc root pass

Arc voltage
12 V
Welding current
150 A
Travel speed
5 in/min
Arc efficiency
0.65
  1. Gross arc energy: 12 x 150 x 60 = 108,000 joules per minute.
  2. Divide by 1,000 x 5 = 5,000 to get 21.6 kJ/in gross.
  3. Multiply by the low open-arc efficiency: 21.6 x 0.65.

Result: 21.6 kJ/in gross, 14.04 kJ/in effective
More than a third of the arc energy is lost. The slow travel speed makes the gross number look substantial, but the plate experiences barely fourteen kilojoules per inch, which is why open tungsten arc roots cool faster than their gross figures suggest.

In practice

  • Efficiency factors are ranges, not constants. Joint geometry, shielding gas, and torch angle all move them, so treat a single published factor as a central estimate rather than a measurement.
  • Most procedure documents specify gross heat input because it is what a welder can compute at the machine. Efficiency belongs in the engineering behind the limit, not in the number the welder is handed.
  • When comparing two processes on the same steel, compare effective heat inputs. Comparing gross figures across processes is close to meaningless.
  • Higher efficiency is not automatically better. It concentrates energy in the joint, which slows cooling and can coarsen grain on steels that would rather cool quickly.

Related topics

Practise the arithmetic under pressure

The welding-math domain in the question bank is where these formulas get tested the way they get tested on paper — with distractors built from the mistakes people actually make.

Welding math questions