How much CFM does my room need?
Type the size of a room and the air changes per hour you want. The calculator gives the airflow in CFM, m³/h and L/s, or works out the air changes from a fan’s CFM.
- Airflow (CFM)
- 120
A 1,440 ft³ room needs 120 CFM for 5 air changes per hour.
- Airflow (m³/h)m³/h
- 203.881
- Airflow (L/s)L/s
- 56.6337
- Air changes per hourACH
- 5
- Room volume
- 1,440 ft³
- Minutes to remove 99% of airborne particlesmin
- 55.3
Airflow (CFM): 120. A 1,440 ft³ room needs 120 CFM for 5 air changes per hour.
How to calculate
Works out the airflow in cubic feet per minute (CFM) a room needs for a number of air changes per hour (ACH), or the air changes a fan’s CFM gives, from the room’s length, width and height.
Example with the default inputs (Find CFM needed, Room length 15 ft, Room width 12 ft, Ceiling height 8 ft, Air changes per hour (ACH) 5): A 1,440 ft³ room needs 120 CFM for 5 air changes per hour.
Method: Volume = length × width × height (ft³); CFM = volume × ACH ÷ 60; ACH = CFM × 60 ÷ volume; minutes to remove 99% = ln(100) ÷ ACH × 60.
- The room is a box: length × width × height. Subtract large built-in volumes yourself.
- ACH counts clean air (outdoor air or filtered air), and the air in the room is well mixed.
- 1 ft = 0.3048 m exactly, so 1 CFM = 1.699 m³/h = 0.4719 L/s.
Worked examples
Each example is checked against the calculator on every build.
- Find CFM needed, Room length 15 ft, Room width 12 ft, Ceiling height 8 ft, Air changes per hour (ACH) 5 gives Airflow (CFM) 120, Air changes per hour 5, Minutes to remove 99% of airborne particles 55.262042.Source: CDC NIOSH, Aim for 5 or More Air Changes per Hour (ACH = CFM × 60 ÷ (area × height)). https://www.cdc.gov/niosh/ventilation/prevention/Aim-for-5.html, retrieved 2026-10-02
- Find Air changes per hour, Room length 20 ft, Room width 15 ft, Ceiling height 10 ft, Airflow (CFM) 300 gives Air changes per hour 6, Airflow (CFM) 300, Minutes to remove 99% of airborne particles 46.051702.Source: Ln(100) ÷ 6 × 60 = 46 minutes, as in CDC, Guidelines for Environmental Infection Control in Health-Care Facilities, Appendix B: Air (ACH = Q ÷ V; Table B.1: 6 ACH removes 99% in 46 minutes). https://www.cdc.gov/infection-control/hcp/environmental-control/appendix-b-air.html, retrieved 2026-10-02
- Find CFM needed, Room length 16.4 ft, Room width 13.12 ft, Ceiling height 8.202 ft, Air changes per hour (ACH) 12 gives Airflow (m³/h) 600, Airflow (CFM) 353.146667, Airflow (L/s) 166.666667.Source: CDC, Guidelines for Environmental Infection Control in Health-Care Facilities, Appendix B: Air (ACH = Q ÷ V; Table B.1: 6 ACH removes 99% in 46 minutes). https://www.cdc.gov/infection-control/hcp/environmental-control/appendix-b-air.html, retrieved 2026-10-02
How it works
- Room volume V = length × width × height, in cubic feet (ft³).
- CFM needed = V × ACH ÷ 60, because an air change every hour moves the whole volume once in 60 minutes.
- Air changes per hour from a known airflow: ACH = CFM × 60 ÷ V.
- The airflow is also shown in m³/h (CFM × 0.3048³ × 60) and L/s (CFM × 0.3048³ × 1,000 ÷ 60).
- Minutes to remove 99% of an airborne contaminant in a well-mixed room with no new source: t = ln(100) ÷ ACH × 60. This is CDC’s removal-time equation, −ln(C₂ ÷ C₁) ÷ ACH × 60, with C₂ ÷ C₁ = 0.01.
Rules
- Each room side is more than 0 and at most 10,000 m; ACH is more than 0 and at most 1,000; CFM is more than 0 and at most 10⁹.
- Lengths can be typed in ft, in, yd, m or cm; 1 ft = 0.3048 m exactly.
- The room is treated as a box with well-mixed air; only clean air counts toward ACH.
Output format. CFM, m³/h, L/s, ACH and the volume to 6 significant figures; the clearing time to 0.1 minute. In Metric the headline is the airflow in m³/h.
Worked examples by hand
15 × 12 ft room, 8 ft ceiling, 5 ACH. V = 15 × 12 × 8 = 1,440 ft³. CFM = 1,440 × 5 ÷ 60 = 120 CFM. Clearing 99%: ln(100) ÷ 5 × 60 = 4.60517 ÷ 5 × 60 = 55.3 minutes.
20 × 15 ft room, 10 ft ceiling, 300 CFM. V = 3,000 ft³. ACH = 300 × 60 ÷ 3,000 = 6. Clearing 99%: 4.60517 ÷ 6 × 60 = 46.1 minutes (CDC Table B.1: 46).
5 × 4 m room, 2.5 m ceiling, 12 ACH. V = 50 m³. Airflow = 50 × 12 = 600 m³/h = 600 ÷ 3.6 = 166.67 L/s. In CFM: 600 ÷ (0.3048³ × 60) = 353.15 CFM.
Other questions people ask
How do I calculate CFM for a room?
Multiply the room’s volume in cubic feet by the air changes per hour, then divide by 60. A 15 × 12 ft room with an 8 ft ceiling holds 1,440 ft³; for 5 air changes an hour it needs 1,440 × 5 ÷ 60 = 120 CFM.
How do I calculate air changes per hour from CFM?
ACH = CFM × 60 ÷ room volume in ft³. A 300 CFM air cleaner in a 20 × 15 ft room with a 10 ft ceiling (3,000 ft³) gives 300 × 60 ÷ 3,000 = 6 air changes per hour.
How many air changes per hour should a room have?
CDC suggests aiming for 5 or more air changes per hour of clean air in rooms people share. Health-care rooms have their own minimums, for example 12 ACH for airborne infection isolation rooms and 6 ACH for standard patient rooms.
How long does it take to clear the air?
In a well-mixed room with no new source, removing 99% of airborne particles takes ln(100) ÷ ACH hours. At 6 ACH that is about 46 minutes, the figure in CDC’s Table B.1.
Do fans that only move air around count?
No. Air changes count clean air: outdoor air from ventilation, or air passed through a filter. A ceiling fan mixes the air but adds no clean air. If you have several sources, add their clean-air CFM together.
How do I convert CFM to m³/h or L/s?
1 CFM = 1.699 m³/h = 0.4719 L/s, from 1 ft = 0.3048 m exactly. The page shows all three.