Residential Duct Leakage CFM25 (IECC R403.3.5)
Whether a residential duct system passes the energy-code tightness test.
Example
You enter
- Measured total leakage at 25 Pa (CFM25) 80
- Conditioned floor area (ft²) 2000
- Limit (CFM25 per 100 ft², default 4) 4
You get
- Normalized leakage 4.00 CFM25 / 100 ft^2
Details, formula, and sources
The measured total leakage at 25 Pa (from a duct blaster) normalized to the conditioned floor area, CFM25 per 100 ft^2 = leakage / area x 100, against the IECC limit (4 total / post-construction, 3 for a rough-in without the air handler). 80 CFM25 on 2,000 ft^2 gives exactly 4.0 -- a pass at the limit; a tighter 60 CFM25 gives 3.0, a leakier 100 gives 5.0 (fail). Distinct from the SMACNA leakage-class test for commercial duct. A field aid; the adopted energy code, the required test type, and the rater govern.
normalized = leakage_cfm25 / cfa_ft2 x 100; passes = normalized <= limit (default 4 CFM25 per 100 ft^2).
IECC (International Energy Conservation Code) §R403.3.5 (duct testing) and RESNET / ANSI 380 duct-leakage test methods, by name.
IECC is viewable through the ICC public-access reader; the leakage / area x 100 normalization is public arithmetic.
Estimate. AHJ and licensed mechanical contractor govern. ACCA Manual J / D / S supersede rules of thumb.
Field names used by the API: leakage_cfm25, cfa_ft2, limit, normalized
- Normalization CFM25 per 100 ft^2 = leakage / conditioned floor area x 100IECC §R403.3.5
- Limit editable 4 total / post-construction, 3 rough-in without the air handler; the edition and test type set itIECC / RESNET
- Rater governs the required test type and the rater's calibrated duct tester govern the official resultscope of this tile