What is The Physics of Swamp Coolers (Evaporative CFM)?
Mathematical Foundation
Laws & Principles
- THE WINDOW RULE (EXHAUST): A swamp cooler does not work in a sealed house. It is a giant air pump. To push 6,000 CFM of cold wet air into the house, 6,000 CFM of hot stale air must exit the house. You must provide 2 square feet of open window space for every 1,000 CFM the cooler creates. If you do not, the house pressurizes, airflow immediately stalls, and the house turns into a hot sauna.
- THE HUMIDITY CEILING: Evaporative coolers strictly rely on dry heat. If the outdoor Relative Humidity exceeds 40%, the water on the pads physically cannot flash-evaporate. The air absorbs no latent heat, the sensible temperature does not drop, and the machine simply becomes a loud fan blowing damp hot air. They are useless in Florida and mandatory in Arizona.
- EXCHANGE RATES BY CLIMATE: In extreme deserts like Phoenix or Death Valley, the sun beats heat through the roof so aggressively that you must replace the entire volume of air inside the house every 2 minutes (T = 2). In a high desert like Reno or a moderate valley, every 3 minutes (T = 3) is sufficient.
Step-by-Step Example Walkthrough
" A homeowner in a scorching Arizona desert wants to install a roof-mounted swamp cooler. The house is 1,800 square feet with standard 8-foot flat ceilings. "
- 1. Find Total Volume: 1,800 sq ft × 8 ft ceiling = 14,400 Cubic Feet of total air mass.
- 2. Identify Exchange Limit: Because it is a hot desert, the air must completely turn over every 2 minutes.
- 3. Calculate CFM: 14,400 Total Volume ÷ 2 Minutes = 7,200 CFM.
- 4. Calculate Exhaust Requirement: 7.2 (Thousands of CFM) × 2 sq ft = 14.4 square feet of open windows required throughout the house.