Pressure decay leak rate

A pressure change test measures the whole system at once: seal it at a known pressure, hold, and see how much pressure is lost. For a fixed volume the leak rate is the pV throughput Q = ΔP·V/Δt, so a 40 mbar loss from a 50 litre system over 10 minutes is 40 × 50 / 600 = 3.3 mbar·L/s. The method is cheap and needs no tracer gas, but it is blind to leak location and it is entirely at the mercy of temperature.

Temperature is the reason most pressure decay tests are wrong. Gas in a rigid volume obeys P/T = constant, so a fall of just 1 K from 293 K drops the pressure by about 0.34% — on a 6 bar absolute system that is 20 mbar of apparent loss with no leak at all. The correction is to refer the final pressure back to the starting temperature, P₂' = P₂ · T₁/T₂, using absolute pressures and absolute temperatures, and to take the leak-induced drop as P₁ − P₂'. Gauge pressures cannot be used in this correction. Where the correction is a large fraction of the raw drop, the test is not really measuring a leak; extend the hold, insulate the item, or move to a tracer-gas method.

The floor on sensitivity is set by the instrument, not the physics: the smallest leak the test can see is roughly the pressure resolution times the volume divided by the hold time. Buying sensitivity therefore means holding longer or measuring pressure better — increasing the volume makes it worse. Allow a thermal stabilisation period before starting the clock, since compression during pressurisation warms the gas and the subsequent cool-down looks exactly like a leak.

Worked example

Test volume50 L
Start pressure (absolute)6000 mbar
End pressure (absolute)5940 mbar
Hold time600 s
Gas temperature at start20 C
Gas temperature at end19 C
Pressure gauge resolution1 mbar
Allowable leak rate1 mbar.L/s
End pressure corrected to start temperature5960.33 mbar
Leak-induced pressure drop39.668 mbar
Drop explained by temperature20.33 mbar
Leak rate3.306 mbar.L/s
Uncorrected leak rate5 mbar.L/s
Smallest detectable leak rate0.0833 mbar.L/s

T1 = 293.15 K, T2 = 292.15 K. P2' = 5940 x 293.15/292.15 = 5960.332 mbar, so 20.33 mbar of the 60 mbar drop was only the gas cooling. The leak-induced drop is 6000 − 5960.332 = 39.668 mbar, giving Q = 39.668 x 50 / 600 = 3.306 mbar·L/s against 5.000 mbar·L/s if temperature were ignored. Detection floor = 1 x 50/600 = 0.0833 mbar·L/s.

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