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§2.4 · Making the weld

Welding-related calculations on the CWI

The CWI calculations area is shop math done by hand: heat input, fillet throat and cross-section, weld metal weight, deposition and cost, and unit conversions. The only help allowed is a four-function, construction or non-programmable scientific calculator (AWS CWI Examination User Guide, September 2026, checked October 2026).

Exam
CWI
Clause
§2.4
Tags here
15
Part A minimum
at least 6%

2.4.1Formulas to carry in your head

The working set for Part A math, customary units
QuantityFormulaUnits to watch
Heat inputvolts × amps × 60 ÷ travel speedJ/in with speed in in/min; ÷ 1,000 for kJ/in
Fillet throat0.707 × leg (equal-leg, flat face)Same unit as the leg
Fillet cross-sectionleg × leg ÷ 2 (equal-leg, flat face)in²; reinforcement not counted
Weld metal weightarea × length × densityLength in inches if area is in in²
Deposition rateweight deposited ÷ arc timelb/h; count arc time only
Operating factorarc time ÷ total timePercent of the shift the arc is on
Deposition efficiencyweight deposited ÷ weight of electrode usedPercent; stubs and spatter are the loss
Length1 in = 25.4 mmExact; round only at the end
Temperature°C = (°F − 32) × 5 ÷ 9Subtract 32 first
Stress1 ksi ≈ 6.895 MPa70 ksi ≈ 483 MPa

Effective throat for strength can differ from the theoretical throat. The code you test on decides which one applies.

2.4.2What the math covers

The questions are short, but Part A is closed book with nothing on screen (AWS Part A information sheet, March 2024, checked October 2026), so every formula and conversion factor has to come from memory.

Calculators are inspected at check-in. Programmable, memory, alphabetical and noisy models aren't allowed. Bring a plain one you already know your way around. Parts A and C are taken at Prometric test centers with no online proctoring (AWS CWI page, checked October 2026); the rest of the setup is on CWI exam format.

Check units before arithmetic: travel speed per minute vs per second, joules vs kilojoules, inches vs millimeters, Fahrenheit vs Celsius. Heat input is where units bite hardest, so it has its own page, heat input.

Fractions on customary drawings

Customary drawings give sizes in fractions of an inch. Know the sixteenths as decimals without reaching for the calculator: 1/16 = 0.0625, 1/8 = 0.125, 3/16 = 0.1875, 1/4 = 0.25, 5/16 = 0.3125, 3/8 = 0.375. Convert to decimals first, then multiply.

Going metric and back

Conversions show up inside bigger questions. A 3/16 in root opening on a hypothetical drawing is 0.1875 × 25.4 = 4.76 mm. A 120°C limit on a hypothetical metric WPS is 120 × 9 ÷ 5 + 32 = 248°F. Do the conversion as its own line and the rest of the problem stays clean.

Two traps hide here. A metric WPS may give travel speed in mm/s. In that case the 60 drops out, and volts × amps ÷ speed gives J/mm directly. And a temperature difference converts without the 32: a 90°F rise is a 50°C rise, while 90°F as a reading is about 32°C.

Cost problems

Cost questions stack several small steps: labor and overhead per hour, how much weld metal goes down per hour of arc, the part of the shift the arc is actually on, and the filler itself. Write each piece with its unit, and the units cancel to dollars per pound or per foot. If they don't cancel, a step is missing.

2.4.3Worked through: heat input in kJ/mm

  1. Write down what you have

    A hypothetical pass at 25 V and 180 A, traveling 8 in/min. The WPS states its limit in kJ/mm.

  2. Volts times amps

    25 × 180 = 4,500 W. A watt is a joule per second.

  3. Seconds to minutes

    4,500 × 60 = 270,000 J per minute of arc time.

  4. Divide by travel speed

    270,000 ÷ 8 = 33,750 J/in, which is 33.75 kJ/in.

  5. Switch to the WPS's unit

    33.75 ÷ 25.4 = 1.33 kJ/mm. Compare that number against the limit.

  6. Sanity-check the size

    Leave out the 60 and you get 562.5 J/in, sixty times too small. That much off is a units slip; no real procedure runs that cold.

2.4.4Pencil math

Calculator out, units written down, then answer. The remarks show the working for every option, wrong ones included.

0 of 15 tagged · 0 accepted

  1. Tag 01

    Calculate the required argon gas flow rate (CFH) for a 1/8 inch tungsten electrode if the rule is 10 CFH per 1/16 inch of electrode diameter

    Remarks on every option
    1. A 25 CFH would be 2.5 steps of 10. A 1/8 in tungsten is exactly two sixteenths.
    2. B 15 CFH is 1.5 steps. 1/8 in = 2/16 in, so it's 2 x 10.
    3. C Correct: 1/8 in = 2/16 in, so 2 x 10 CFH = 20 CFH.
    4. D 10 CFH covers 1/16 in only. A 1/8 in tungsten is twice that.

    Pick an option. The remarks on all 4 open here.

  2. Tag 02

    Calculate the required filler metal weight for a 20-foot long V-groove weld if the deposition rate is 0.25 pounds per foot

    Remarks on every option
    1. A Correct: 20 ft x 0.25 lb/ft = 5 lb.
    2. B 4.5 lb would be 18 ft at 0.25 lb/ft. Use the full 20 ft.
    3. C 4 lb would be 16 ft. Use the full 20 ft.
    4. D 5.5 lb overshoots. 20 x 0.25 is exactly 5.

    Pick an option. The remarks on all 4 open here.

  3. Tag 03

    Calculate the required preheat time in minutes to reach 400°F if the heating rate is 15°F per minute starting from 70°F

    Remarks on every option
    1. A 25 min is 375°F of rise. You only need 400 - 70 = 330°F.
    2. B 24 min is 360°F of rise, too much. The rise is 330°F.
    3. C 20 min is only 300°F, which leaves you at 370°F.
    4. D Correct: 400 - 70 = 330°F to gain; 330 ÷ 15 = 22 min.

    Pick an option. The remarks on all 4 open here.

  4. Tag 04

    Calculate the maximum allowable amperage for a 3/32 inch tungsten electrode if the guideline is 1,600 amps per inch of electrode diameter.

    Remarks on every option
    1. A 145 A comes from rounding 3/32 down. 3/32 in is exactly 0.09375 in.
    2. B 140 A is short. 0.09375 x 1,600 = 150 A.
    3. C 160 A would be a 0.1 in tungsten. 3/32 in is a bit smaller.
    4. D Correct: 3/32 in = 0.09375 in; 0.09375 x 1,600 = 150 A.

    Pick an option. The remarks on all 4 open here.

  5. Tag 05

    If welding cost is $75/hour and deposition rate is 5 lbs/hr, what is the cost per pound of deposited weld metal?

    Remarks on every option
    1. A Correct: $75 per hour ÷ 5 lb per hour = $15 per lb.
    2. B $18 doesn't come out of these numbers. $75 ÷ 5 is $15.
    3. C $20 would mean $100 an hour at 5 lb/h.
    4. D $12 would mean $60 an hour, or 6.25 lb/h.

    Pick an option. The remarks on all 4 open here.

  6. Tag 06

    If welding operation costs are $85/hour and materials cost $25/lb, what is the total cost to deposit 4 lbs of weld metal at a deposition rate of 2 lbs/hour?

    Remarks on every option
    1. A $255 is three hours of labor alone. You need 2 h of labor ($170) plus $100 of material.
    2. B Correct: 4 lb ÷ 2 lb/h = 2 h, so labor is 2 x $85 = $170. Material is 4 x $25 = $100. Total $270.
    3. C $240 is short. Labor is $170 and material $100.
    4. D $285 overcounts. $170 labor + $100 material = $270.

    Pick an option. The remarks on all 4 open here.

  7. Tag 07

    What is the electrode efficiency if 8 pounds of electrode produces 6.4 pounds of deposited weld metal?

    Remarks on every option
    1. A 85% would mean 6.8 lb deposited.
    2. B Correct: 6.4 ÷ 8 = 0.80, so 80% of the electrode ended up as weld metal.
    3. C 75% would mean 6.0 lb deposited.
    4. D 70% would mean 5.6 lb deposited.

    Pick an option. The remarks on all 4 open here.

  8. Tag 08

    If a welder deposits 5 pounds of weld metal in 2 hours, what is the deposition rate in pounds per hour?

    Remarks on every option
    1. A Correct: 5 lb ÷ 2 h = 2.5 lb/h.
    2. B 5 lb/h is the total, not divided over the 2 hours.
    3. C 3 lb/h would be 6 lb in 2 hours.
    4. D 2 lb/h would be 4 lb in 2 hours.

    Pick an option. The remarks on all 4 open here.

  9. Tag 09

    If the welding arc time is 6 minutes out of a 10-minute period, what is the duty cycle?

    Remarks on every option
    1. A 55% would be 5.5 min of arc time.
    2. B 65% would be 6.5 min of arc time.
    3. C 50% would be 5 min of arc time.
    4. D Correct: 6 ÷ 10 = 60%. Duty cycle is figured on a 10-minute period.

    Pick an option. The remarks on all 4 open here.

  10. Tag 10

    Calculate the included angle needed for a V-groove weld if each bevel angle is 35 degrees

    Remarks on every option
    1. A 60° would mean 30° bevels.
    2. B 80° would mean 40° bevels.
    3. C 75° would mean 37.5° bevels.
    4. D Correct: The included angle is both bevels added: 35° + 35° = 70°.

    Pick an option. The remarks on all 4 open here.

  11. Tag 11

    Calculate the required gas flow rate in L/min if the current specification is 35 CFH (Cubic Feet per Hour)

    Remarks on every option
    1. A 15.75 L/min is too low. One cubic foot is about 28.3 L, so 35 CFH is about 991 L per hour; divide that by 60.
    2. B Correct: 35 ft³/h × 28.32 L/ft³ ≈ 991 L/h, and 991 ÷ 60 ≈ 16.52 L/min.
    3. C 17.25 L/min is too high. Use 28.32 L per cubic foot and divide the hourly volume by 60 minutes.
    4. D 18.00 L/min overshoots. With 28.32 L per cubic foot, 35 CFH works out to about 16.5 L/min.

    Pick an option. The remarks on all 4 open here.

  12. Tag 12

    Calculate the required gas volume in cubic feet for a 4-hour welding shift if flow rate is 40 CFH and arc time is 65%

    Remarks on every option
    1. A 100 cu ft is about 2.5 h of arc. 65% of 4 h is 2.6 h.
    2. B Correct: Arc time is 4 h x 0.65 = 2.6 h; 2.6 x 40 CFH = 104 cu ft.
    3. C 110 cu ft is too high. 2.6 h x 40 = 104.
    4. D 95 cu ft is short. Gas flows during arc time: 2.6 h x 40 = 104.

    Pick an option. The remarks on all 4 open here.

  13. Tag 13

    Calculate the overlap percentage if the weld bead width is 0.5 inches and the step increment is 0.3 inches

    Remarks on every option
    1. A Correct: Each bead overlaps the last by 0.5 - 0.3 = 0.2 in; 0.2 ÷ 0.5 = 40%.
    2. B 45% would need a 0.275 in step.
    3. C 50% would need a 0.25 in step, half the bead width.
    4. D 35% would need a 0.325 in step.

    Pick an option. The remarks on all 4 open here.

  14. Tag 14

    If a welding machine uses 35 volts and 175 amps, what is the power consumption in kilowatts?

    Remarks on every option
    1. A Correct: 35 V x 175 A = 6,125 W = 6.125 kW.
    2. B 6.5 kW is too high. 35 x 175 = 6,125 W.
    3. C 6.25 kW is 125 W too high. 35 x 175 = 6,125 W.
    4. D 5.875 kW is 250 W short. 35 x 175 = 6,125 W.

    Pick an option. The remarks on all 4 open here.

  15. Tag 15

    Calculate the weld metal weight in pounds needed for a 30-foot long, 1/4 inch fillet weld if the density is 0.28 lbs/cubic inch

    Remarks on every option
    1. A 3.45 lb overshoots. A 1/4 in fillet's cross-section is ½ x 0.25 x 0.25 = 0.03125 sq in.
    2. B 3.75 lb is too much. 0.03125 sq in x 360 in = 11.25 cu in.
    3. C 2.85 lb is short. 11.25 cu in x 0.28 = 3.15 lb.
    4. D Correct: Area = ½ x 0.25² = 0.03125 sq in; x 360 in = 11.25 cu in; x 0.28 lb/cu in = 3.15 lb.

    Pick an option. The remarks on all 4 open here.

2.4.5Sources

  1. AWS CWI Examination User Guide, September 2026 — calculators allowed in the test room (checked October 2026)
  2. AWS Part A information sheet, March 2024 — Part A is closed book (checked October 2026)
  3. AWS CWI page — Parts A and C at Prometric centers (checked October 2026)

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