How this calculation works
A duct leakage reading depends on the pressure applied during the test. Comparing raw readings at different pressures can confuse a change in test conditions with a change in leakage paths. A power-law conversion puts the values on a stated pressure basis.
Reference leakage = measured leakage × (reference pressure/test pressure)^exponentA worked example
A 100 L/s test at 50 Pa converted to 25 Pa with exponent 0.65 becomes about 63.728 L/s. The pressure is halved, but leakage is not halved because the exponent is below one. For 100 m² of duct surface, that is about 0.63728 L/s per square metre. Relative to an entered 1000 L/s system flow, it is about 6.373 percent on the chosen pressure basis.
Good to know
Fixed-path power law only; no test calibration, standard leakage class, whole-network pressure solve or energy savings estimate. Changing the pressure exponent can change the comparison materially. Use the relevant measured or specified exponent and test procedure rather than assuming the example default is universally applicable.
Reference: PNNL — duct leakage power-law conversion (opens in a new tab) ↗
What the model answers
A duct leakage reading depends on the pressure applied during the test. Comparing raw readings at different pressures can confuse a change in test conditions with a change in leakage paths. A power-law conversion puts the values on a stated pressure basis.
The exponent describes how flow through the leakage paths responds to pressure. It is supplied here and held fixed; one test point cannot independently identify both that exponent and a leakage coefficient.
The Ventilation air changes and dilution time calculator needs effective clean-air delivery to the room. A normalized duct leakage result alone does not establish that delivery rate.
Measure and enter consistently
Use positive pressure magnitudes for both test and reference conditions. The equation does not use the sign of a supply or extract pressure. Flow units are litres per second in both the measured and comparison values.
Duct surface area is external duct-envelope area for the normalized area metric, not duct cross-sectional flow area. Entered system flow provides a separate percentage comparison and does not modify the leakage conversion.
Interpret the result
A lower reference pressure produces lower normalized flow for a positive exponent. An equal reference and test pressure must reproduce the measured reading exactly, which is a useful unit and input check.
The per-area and system-flow ratios are arithmetic comparisons. They do not assign a leakage class, establish a pass/fail threshold or predict leakage throughout a network with varying pressure and different leakage locations.
Work through the numbers
A 100 L/s test at 50 Pa converted to 25 Pa with exponent 0.65 becomes about 63.728 L/s. The pressure is halved, but leakage is not halved because the exponent is below one.
For 100 m² of duct surface, that is about 0.63728 L/s per square metre. Relative to an entered 1000 L/s system flow, it is about 6.373 percent on the chosen pressure basis.
Where this model stops
Fixed-path power law only; no test calibration, standard leakage class, whole-network pressure solve or energy savings estimate.
Changing the pressure exponent can change the comparison materially. Use the relevant measured or specified exponent and test procedure rather than assuming the example default is universally applicable.
A few questions, answered
Why not scale leakage directly with pressure?
Direct proportionality assumes exponent one. A different measured or specified exponent gives a different pressure response.
Can this certify a duct leakage class?
No. The page has no jurisdiction or standard-specific acceptance thresholds.