Welding
Argon purge time calculator
In short
A back purge shields the root side of a weld. The volume between dams is V = π/4 × Di² × L; the gas to pass before welding is n × V and the pre-purge time t = n × V / Q. A perfect mixing model, C(t) = C0 · e^(−Q·t / V), also gives the time to reach an oxygen limit.
Calculation
Example calculated with the tool’s default values.
Inputs
- Volume
- Pipe between two dams
- Pipe
- DN50 · 60.3 × 2.9 mm
- Purged length between dams
- 300 mm
- Material
- Austenitic stainless 316L
- Purge gas
- Argon
- Volume changes
- 5
- Pre-purge flow
- 15 L/min
- Welding time under purge
- 10 min
- Limit to reach at the vent
- By material
Results
- Minimum pre-purge time
- 21 s100 ppm limit with perfect mixing
Worked example
Every value in this section belongs to the example, calculated with the tool’s default values.
Other results of the example
- Pre-purge time
- 14 s5 volume changes
- Time to reach the limit
- 21 s100 ppm, perfect mixing
- Volume changes for the limit
- 7.6
- Purged volume
- 0.70 L
- Pre-purge gas
- 5.4 L
- Gas while welding
- 50 L5 L/min (manufacturer)
- Total gas use
- 55 L
- Inside diameter
- 54.5 mm
- Target residual oxygen
- 100ppm maximum: 100 (BOC), 50 at Huntingdon, 20 for severe service; WPS takes priority
- Indicative manufacturer-practice values: they do not replace the qualified WPS or an oxygen reading.
- Only an oxygen analyser inside the purged zone confirms the purge; time alone is not enough.
- Provide an exit vent and, while welding, a barely perceptible flow at the exit (Huntingdon): too much flow builds up pressure in the tube.
- Large volumes of purge gas: asphyxiation risk around the station, check ventilation (BOC).
- Perfect mixing (worst case): 1415 ppm O2 after 5 volume changes, above 100 ppm. Bottom-to-top displacement does better in practice: check with an analyser.
- The 100 ppm limit needs 7.6 volume changes with perfect mixing: pre-purge raised to 21 s.
Step-by-step calculation
Inside diameter
Di = OD − 2e
60.3 − 2 × 2.90 = 54.5 mm
Purged volume
V = π/4 × Di² × L
π/4 × 54.5² × 300 = 699849 mm³ = 0.700 L
Pre-purge gas
Vg = Q × t chambrage
15.0 L/min × 21 s / 60 = 5.35 L
Pre-purge time
t = n × V / Q
3.50 / 15.0 = 0.23 min = 14 s
Oxygen, perfect mixing (worst case)
C = 210 000 ppm × e^(−n)
210 000 × e^(−5.0) = 1415 ppm
Volume changes for 100 ppm (perfect mixing)
n = ln(210 000 / 100)
7.6
Oxygen at the vent
C(t) = C0 · e^(−Q·t / V)
C0 = 210 000 ppm ; Q = 15.0 L/min ; V = 0.700 L
Time for 100 ppm
t = V / Q × ln(C0 / C)
0.700 / 15.0 × ln(210 000 / 100) = 0.36 min = 21 s
Minimum pre-purge
max(t renouvellements ; t seuil)
max(14 ; 21) = 21 s
How to use it
- Pick the pipe (outside diameter and wall thickness) and the purged length between dams, or enter a known volume.
- Pick the material and the purge gas: argon or nitrogen.
- Enter the number of volume changes, the pre-purge flow and the welding time under purge.
- Pick the oxygen limit to reach at the vent, or enter the WPS value in ppm.
- Read the minimum pre-purge time.
Frequently asked questions
How do I calculate the argon purge time of a pipe?
Example: a 60.3 × 2.9 mm pipe, so Di = 54.5 mm, over 300 mm between dams: V = π/4 × 54.5² × 300 = 0.700 L. With 5 volume changes at 15 L/min, t = n × V / Q = 3.50 / 15.0 = 0.23 min, or 14 s. To reach 100 ppm of oxygen with perfect mixing, t = V / Q × ln(C0 / C) = 21 s, so the minimum pre-purge time is 21 s.
How is back purge welding done?
Close the pipe on both sides of the joint (dams, bladders or soluble film), provide an exit vent and replace the air with purge gas before welding. The manufacturer practice used is about 5 volume changes at 15 to 20 L/min, then about 5 L/min while welding, barely perceptible at the exit.
Which gas for a back purge on stainless?
Argon suits all materials. Oxygen-free nitrogen is allowed by BOC on austenitic and duplex grades, but it stabilises austenite and can give an unacceptable microstructure: the calculation keeps it to non-critical work on austenitic (excluded in ASME BPE). On duplex the sources disagree: follow the WPS, otherwise argon.
What oxygen level should the purge reach?
On stainless, 100 ppm (0.01 %) is considered satisfactory by BOC, 50 ppm at Huntingdon, 20 ppm for severe service. On carbon steel, purging is normally unnecessary; when required, 1 % O2 is often considered acceptable. The WPS takes priority, and only an oxygen analyser confirms the purge.
How many volume changes for a duplex purge?
IMOA recommends pure argon fed from the bottom and at least 7 volume changes; the calculation warns below that. With perfect mixing, 5 volume changes still leave 1415 ppm of oxygen: displacement purging does better in practice, but check with an analyser.
Limits
Indicative results, taken from manufacturer practice (Huntingdon, BOC) and IMOA: they replace neither the qualified WPS nor an oxygen reading. The perfect mixing model is the worst case; only an analyser inside the purged zone confirms the purge. A flow above 20 L/min creates turbulence that lengthens the purge. Large volumes of purge gas: asphyxiation risk, check ventilation.
Sources
- Huntingdon Fusion Techniques : pré-purge (≈ 5 renouvellements, 15–20 L/min, 5 L/min en soudage) : Shielding Gas and Purging Techniques during Welding, Part 2 (link)
- Huntingdon Fusion Techniques : oxygène résiduel inox < 50 ppm : Welding Stainless Steel (link)
- BOC / Linde : gaz de purge, teneurs en O2 admises, azote : The right gas working for you, Purging or root backing (link)
- Orbital Welding Guide : azote, travaux non critiques: Argon Purge Welding Guide (link)
- Dilution : équation de purge d’un volume (mélange parfait) : Basic room purge equation, t = V/Q · ln(C0/C) (link)
- IMOA : envers à l’argon, par le bas, au moins 7 fois le volume : Practical Guidelines for the Fabrication of Duplex Stainless Steels, § 12.3.1 (link)
- Géométrie : Volume d’un cylindre