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[Chris Parks]

I will put this here and if it is in the wrong place will the mods please move it. Some time ago there was a post relating to how long it takes an exhaust fan or dust extractor to change over the air in a room. It explained that the cubic capacity of the fan to move air does not necessarily mean it will cause a complete change as a direct mathematical relationship would suggest. My apologies if this is a bit vague but it is the best I think I can do. I have got no idea on how to search for it as it was just one post in a thread and I hope either the author can point me in the right direction or someone recalls it. Thanks.

[Tom Veatch]

I don't recall seeing that post, but if by "complete change over" you mean how long will it take for an exhaust fan to replace every molecule of air in a room with make up air from outside the room, theoretically the answer is "forever". The incoming air mixes with the air already in the room and the exhaust fan can't distinguish between air that was in the room when it began to run from that air that entered the room after the fan started.

Therefore, shortly after the fan begins to run, the exhaust is a mixture of "old" and "new" air. That's why a "complete" changeover of air in a 1000 cu ft room with a 100 cu ft/minute exhaust fan takes much longer than 10 minutes to scavenge all the "old" air out of the room.

Well, then, how long does it take?

First you have to decide what you mean by "complete". 50% old air remaining? 25%? 10%? etc. Then you have to determine (or assume) how the incoming "new" air mixes with the existing "old" air. The simplest assumption, knowing nothing about the flow dynamics in the room, and good only as a first approximation, is that the amount of old air in the exhaust is the same percentage as the amount of old air still in the room.

With that assumption, a differential equation can be written in which the reduction in old air per unit time is proportional to the instantaneous percentage of old air remaining in the room. The theoretical solution of that equation is asymptotic to zero. That is, as time increases, the volume of old air gets closer and closer to zero but never quite reaches zero in any finite time.

If my remembrance of differential equations hasn't completely deserted me, the solution is:

T/To = ln(1/R)

Where:

T = amount of time to reduce the amount of "old" air to a given percentage
To = Time Constant (length of time to exchange all the old air if there were no mixing = volume of room/flow rate of the fan = 10 minutes for our example case)
R = percentage of "old" air remaining in the room.

Table of Values:
R, T/To, Time for that 1000 cu ft room with the 100 cu ft/min fan
1, 0, 0
.9, .105, 1.05 minutes
.8, .223, 2.23 minutes
.7, .357, 3.57 minutes
.6, .511, 5.11 minutes
.5, .693, 6.93 minutes
.4, .916, 9.16 minutes
.3, 1.204, 12.04 minutes
.2, 1.609, 16.09 minutes
.1, 2.303, 23.03 minutes
.05, 2.996, 29.96 minutes
.01, 4.605, 46.05 minutes

So looking at those values, at the 10 minute mark, when the simple calculation says all the air has been changed, a little less than 40% of the old air still remains in the room. After 20 minutes, when the air should have been completely exchanged twice, about 15% of the old air is still there. Even at the 46 minute mark there's still 1% of the original air still in the room.

So, it takes much longer to exchange the air in a room than a simple volume/exchange rate indicates. Based on those numbers, a rule of thumb might be "To get 95% exchange, triple the simple calculation."