“A 20x20x8 ft room is full of air containing 1000 pollen grains per cubic foot, a HEPA filter runs in the room, stirring the air well. The HEPA filter removes 99.997% of pollen grains that pass through the filter, and moves r=300 cfm of air. How long before the room is down to 1 pollen grain per cubic foot?”

It’s a more or less textbook problem for Calculus and/or differential equations classes, but it’s worth thinking about how to set it up, and also it’s worth knowing, because damn the allergens are serious around here.

Using nonstandard analysis, we’ll suppose we have an infinitesimal time $$dt$$ in which a volume of air $$r dt$$ passes through the filter, with the concentration of pollen in the exhaust being $$0.00003 C$$ and the concentration in the remaining air remaining as $$C$$. The overall concentration is then $$(C(V-r dt)+0.00003 C r dt)/V$$. The Change in concentration per unit time is $$(C(dt)-C(0))/dt$$ which works out to $$dC/dt = -0.99997 r C /V$$, or $$\frac{1}{C}\frac{dC}{dt} = -0.99997 r/V$$ which has solution $$C = C_0e^{-.99997rt/V}$$ (differentiate to verify) since .99997 is about 1, let’s just say we’re looking for $$e^{-rt/V} = 1/1000$$ or $$t = \log(1000)V/r$$ or plugging in our numbers, about 74 minutes for our example room!

So although it only takes 11 minutes or so to move one roomful of air through the filter, because the clean air mixes with the “dirty” it takes 74 minutes to fully scrub the air. Also, the time is proportional to the ratio V/r, so for a fixed size filter, the bigger the room, the longer the time to clean it, scaling linearly. Get a big filter, and run it for a lot longer than you think!