§2.4A · Under Welding calculations
Heat input on the CWI: the formula and what it changes
Heat input is the arc energy that goes into each inch of weld, and on the CWI you build it from three readings: voltage, current and travel speed. Line the units up first and the arithmetic takes one line. The same number decides a lot about the steel next to the weld.
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- CWI
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- §2.4A
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- Welding calculations
2.4A.1Worked example: one vertical fill pass
Collect the readings
The pass is gas-shielded flux core, and the meter shows 25 V and 190 A. The welder covers 15 in in 2 minutes, so travel speed is 15 ÷ 2 = 7.5 in/min.
Volts times amps
25 × 190 = 4,750 W. A watt is a joule per second.
Times 60
4,750 × 60 = 285,000 J every minute, which now matches a speed given per minute.
Divide by travel speed
285,000 ÷ 7.5 = 38,000 J/in.
Convert, then apply efficiency only if asked
38,000 J/in = 38.0 kJ/in. If a question supplies a thermal efficiency factor, multiply last: with 0.8, you'd get 30.4 kJ/in.
2.4A.2Keep the units straight
| Given | Calculate | Answer in |
|---|---|---|
| Speed in in/min | V × A × 60 ÷ speed | J/in |
| Speed in in/s | V × A ÷ speed (no 60) | J/in |
| Speed in mm/min | V × A × 60 ÷ speed | J/mm |
| Weld length and arc time | Speed = length ÷ time, then as above | J/in or J/mm |
| V, A and arc time only | V × A × seconds | J for the whole weld |
1 in = 25.4 mm, so 1 kJ/mm = 25.4 kJ/in. The last row is total energy, not heat input. There's no length in it.
2.4A.3What more heat does to the joint
Heat input is one of three things that set cooling rate, along with preheat and interpass temperature and thickness. Change the heat input and the cooling rate moves with it. The cooling rate in turn moves the heat-affected zone: how wide it gets, what its grain looks like and how tough it ends up.
Cut it too far and the weld cools fast. On a hardenable steel that means a hard HAZ and a higher risk of hydrogen cracking. That's why a WPS can carry a maximum for toughness and, on crack-prone steel, effectively a minimum too. The heat control and metallurgy clause goes deeper on the metallurgy.
Same amps, different heat input
The formula lumps current, voltage and speed into one number, so the welder's hands matter. A wide weave covers the joint slower, and lower travel speed means higher heat input even with the machine settings untouched. Stringer beads keep it down.
Checking it in the field means timing travel over a marked length, reading the meter during welding and comparing both with the WPS ranges. That job sits in welding application and control. The welding calculations clause also covers fillet size and throat.
2.4A.4Volts, amps, inches
Have a calculator out. Write the units next to every number before you multiply, and the setup takes care of itself.
0 of 7 tagged · 0 accepted
Tag 01
What is the heat input (kJ/in) for a weld if the current is 200 amps, voltage is 24V, travel speed is 12 inches per minute, and efficiency factor is 0.85?
Remarks on every option
- A Wrong arithmetic. Arc energy is 24.0 kJ/in, and 0.85 of that is 20.4, not 22.8.
- B Doesn't match. It's above even the unadjusted 24.0 kJ/in.
- C Correct: 24 V × 200 A × 60 ÷ 12 in/min = 24,000 J/in = 24.0 kJ/in. Times 0.85 = 20.4 kJ/in.
- D Too low. 0.85 × 24.0 kJ/in is 20.4.
Pick an option. The remarks on all 4 open here.
Tag 02
How is heat input in welding typically calculated?
Remarks on every option
- A Upside down. Volts times amps goes on top.
- B Multiplying by travel speed gets it backwards. Faster travel lowers heat input.
- C Inverted. Heat input rises with current and voltage.
- D Correct: Heat input = volts × amps ÷ travel speed, times 60 when speed is in in/min, for J/in.
Pick an option. The remarks on all 4 open here.
Tag 03
Calculate the total energy delivered by the arc, in kJ, if voltage is 28 V, current is 225 A and arc time is 2.5 minutes.
Remarks on every option
- A Too high. 28 V × 225 A × 150 s = 945,000 J.
- B Doesn't match. Convert 2.5 min to 150 s first.
- C Doesn't match. 6,300 W × 150 s = 945 kJ.
- D Correct: Power = 28 × 225 = 6,300 W. Over 150 s that's 945,000 J, or 945 kJ.
Pick an option. The remarks on all 4 open here.
Tag 04
Which welding parameter has the greatest effect on cooling rate?
Remarks on every option
- A Correct: Heat input sets how fast the weld cools, along with preheat and thickness. More heat input, slower cooling.
- B Arc length shifts voltage a little, but it affects cooling only through heat input.
- C Electrode size sets the current range. Cooling rate tracks heat input.
- D Joint geometry affects heat flow, but it isn't a welding parameter you set.
Pick an option. The remarks on all 4 open here.
Tag 05
How does heat input affect the size of the heat-affected zone (HAZ)?
Remarks on every option
- A Heat input directly affects how far the heat spreads.
- B Backwards. Less heat input gives a narrower HAZ.
- C Correct: More heat input spreads heat farther into the base metal, widening the HAZ.
- D Every fusion weld has a HAZ.
Pick an option. The remarks on all 4 open here.
Tag 06
During welding of 1-inch thick low-alloy steel plate, what effect would doubling the heat input have on the grain size in the heat-affected zone?
Remarks on every option
- A Doubling heat input changes the thermal cycle, and the HAZ grains respond.
- B Weld metal grains do respond, but so does the HAZ. The coarse-grained zone next to the fusion line grows with heat input.
- C Correct: Doubling heat input keeps the HAZ hot far longer, and the grains keep growing the whole time they sit in the grain-growth range.
- D Backwards. Grain size goes up, not down, with more heat.
Pick an option. The remarks on all 4 open here.
Tag 07
Which condition best indicates excessive heat input during welding when performing visual inspection?
Remarks on every option
- A An even surface points to steady technique, not too much heat.
- B Correct: Heavy scale and heat tint well beyond the weld point to too much heat input.
- C Uniform ripples and smooth tie-in point to steady, controlled parameters.
- D Consistent width and height show steady travel. Excess heat tends to widen and flatten the bead.
Pick an option. The remarks on all 4 open here.
2.4A.5Calculator talk
Do I need the efficiency factor on the CWI?
Only when the question gives you one. Without it, the plain volts × amps × 60 ÷ speed result is the answer being asked for.
Is arc energy the same as heat input?
Depends on who's writing. Many US texts call the uncorrected number heat input. ISO practice calls it arc energy and keeps heat input for the value after the efficiency factor. Volts × amps × arc time is a third thing: total energy for the whole weld. Some questions call that total arc energy too, in kJ, so go by the units the answer asks for: kJ alone or kJ/in.
Does preheat count toward heat input?
No. Heat input covers the arc only. Preheat is a separate number on the WPS, even though both slow the cooling of the joint.
How do I get travel speed if nobody gives it to me?
Mark a length on the joint, time the pass with a stopwatch and divide length by minutes. Ten inches in 80 seconds is 10 ÷ 1.33, about 7.5 in/min.
More tags on your break
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