Calculate purge time and gas consumption for dilution purging of gas lines: t = V/Q × ln(C0/C) - for commissioning high-purity gas systems.
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How long must a gas line be purged?
During dilution purging the contamination falls exponentially: C(t) = C₀ × e^(−Q·t/V). The target is reached after t = V/Q × ln(C₀/C). Practical consequence: each volume exchange reduces contamination by a factor of e ≈ 2.7 – getting from air oxygen (209,000 ppm) down to 1 ppm takes ln(209,000) ≈ 12.3 volume exchanges.
Worked example
New stainless line 12×1 mm (ID 10 mm), 50 m long: volume = π × (0.005 m)² × 50 m = 3.9 litres. At 20 Nl/min nitrogen purge flow: t = 3.9/20 × 12.3 = 2.4 minutes theoretical, about 4 minutes with a 1.5 safety factor, consuming 72 Nl of nitrogen. Real-world purges take longer because of dead volumes – and because moisture adsorbed on tube walls desorbs far more slowly than oxygen dilutes.
Why the formula is not enough for moisture
Water is the most stubborn contaminant: it adsorbs on surfaces and desorbs over hours to days. The dilution formula only gives the absolute lower bound. Proven practice: purge with dry gas (dew point ≤ −70 °C), keep the line warm, monitor with a dew point analyser.
Alternative: pressure-cycle purging
For branched systems and dead legs, pressure cycling is more efficient: fill to p₁, vent to p₂ – each cycle reduces contamination by p₁/p₂. Three cycles 10 → 1 bar equal a factor of 1000.
