Thread by @joeyfox85

1 air change only removes 63% of the virus particles. Why? (Warning: Math)

Thought experiment 1:
Start with 100 virus particles (VPs).
Remove all the air in the room (100 VPs).
Refill the room with air.
You have 0 VPs left - 100% removal.

1/5

Thought experiment 2:
Start with 100 virus particles (VPs).
Remove half the air in the room (50 VPs).
Refill half the room with air and let it mix.
You have 50 VPs left.
Remove half of the air again (25 VPs).
Refill the room.
You have 25 VPs left - 75% removal

2/5

Thought exp. 3:
Start with 27 VPs.

Remove 1/3 of the air (9 VPs).
18 VPs remain. Refill the room.

Remove 1/3 of the air (6 VPs).
12 VPs remain. Refill the room.

Remove 1/3 of the air (4 VPs).
8 VPs remaining, 19 removed total.
Refill the room.

19/27=70% of VPs removed.

3/5

How it's really done: calculus

The rate of virus removal is proportional to the concentration of the virus:
-d(virus removed)/d(volume removed)= C * virus concentration
C is a constant.

This is the classic mixing problem - a differential equation.

4/5

Once you solve it, you get:
Old air remaining %=1/e^(air changes)

Fresh air % = 1-old air%=1-1/e^(air changes)

AC Fresh Air %
1 63
2 86
3 95
4 98
5 99

5/5