What DIN 1946-6 calculates
The standard first decides whether ventilation measures are needed at all for a dwelling, based on floor area, occupancy, insulation and airtightness. When they are, it gives the nominal ventilation flow q_NL, which depends on the usable floor area A of the dwelling:
q_NL = −0.002 × A² + 1.15 × A + 11, in m³/h.
For a dwelling of 201.8 m² this gives 162 m³/h.
Minimum extract per room
The total flow is the higher of q_NL and the sum of the minimum extract rates:
| Room | Minimum extract (m³/h) |
|---|---|
| Kitchen | 45 |
| Bathroom | 45 |
| Toilet | 25 |
| Utility room (HWR) | 25 |
Worked example
A home of 100 m². q_NL = −0.002 × 100² + 1.15 × 100 + 11 = 106 m³/h. The minimum extract of kitchen (45), bathroom (45) and toilet (25) adds up to 115 m³/h. The total is the higher value, 115 m³/h.
How Aira applies it
Aira counts all rooms inside the dwelling as floor area and divides the supply over the rooms with factors per room type: living room 3, bedroom 2, dining, work and guest room 1.5. Reduced (0.7 × q_NL) and intensive (1.3 × q_NL) ventilation are operating modes and are shown below the balance. The tool does not calculate the 30 m³/h per person and the moisture-protection flow with infiltration.
The Lüftungskonzept itself, including whether one is required, is not calculated by Aira. You can adjust every figure in the table before you download the plan.
Source
DIN 1946-6:2019-12 is sold by the standards publisher and is not available free of charge. Aira follows the published descriptions of the standard from German trade publications and industry associations, and offers the country as a closed beta until the figures are compared with the full text.
- DIN 1946-6:2019-12, Lüftung von Wohnungen