The actual conditions were that 2 oz (30 mL) were added to the surface of a pool 20 ft. by 40 ft. (~50 sq. meters), the pool room assumed to be 30 X 50 X 20 or to contain 30,000 cu.ft. (~600 cu.m.) the ventilation system to give one change of air per hour, the water temperature to be 82 degrees F. and the air temperature varying from 85 to 75 F and the surface wind speed of 2 miles per hour.

The data used in the calculations were partly from laboratory and actual pool observations such as the evaporation of the isopropyl alcohol was essentially completed in less than one minute and the rate of spread on a clean surface was about 500 mm/min. on a clean surface and about 250 mm/min (~10 ft/min.) on a pool surface. Using these values and a worst case situation for 2 - 100,000th of an inch layer and should not penetrate the water to a depth of more than a few millionth of an inch even when dumped in rather than drizzled or meter pumped. Diffusion of this depth to the surface and consequent evaporation will constitute the slowest release to the air and should give undetectable amounts (there will always be some in the water as long as there is some in the air) in 10 minutes providing the ventilation system is working. The worst case for the air in the pool room is 1/4 part per million initially rapidly decreasing as the ventilation system changes the air so that at 10 minutes it will be 1/5 to 1/10 ppm and at one hour will be about 10 parts per billion.

Considering the outdoor pool, the water concentration of isopropyl alcohol will be a little less (and continuing less) than the indoor pool since within 2 minutes there will be no detectable isopropyl alcohol in the air and the concentration which would have been 2/10 of a part per million at 1 minute will as noted by undetectable in 2 minutes leaving the water concentration undetectable in 10 minutes or less.

Signed,
Robert N. O'Brien,
Professor Emeritus