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The concentration of a pollution in a limited space as a room, container, tank etc. depends on

The concentration of a pollution can be calculated like
c = q / n V (1 - e-nt) + (c1 - c2) e-nt + c2 (1)
where
c = concentration in the space at perfect mix (m3/m3) (kg/kg)
q = amount of pollution added to the space (m3/h) (kg/h)
n = volume changes per hour (h-1)
V = volume or mass of the space (m3) (kg)
t = time (h)
c1 = concentration in the space at start (m3/m3) (kg/kg)
c2 = concentration in the supply fluid (m3/m3) (kg/kg)
If the start concentration (t = 0) in the space and the concentration in the supply fluid is zero, equation (1) can be reduced to
c = q / n V (1 - e-nt) (2)
Note! After some time the concentration in the room will stabilize. The table below shows the value of c when the amount of pollution q = 1, and the volume of the space V = 1.
| Concentration of pollution - c - (m3/m3) (kg/kg) | |||||||||
| Volume changes per hour
(1/h) |
Time - t - (min) | ||||||||
| 10 | 20 | 40 | 80 | 160 | 320 | 640 | 1280 | 2560 | |
| 0.5 | 0.1599 | 0.3070 | 0.5669 | 0.9732 | 1.4728 | 1.8610 | 1.9903 | 2.0000 | 2.0000 |
| 1 | 0.1535 | 0.2835 | 0.4866 | 0.7364 | 0.9305 | 0.9952 | 1.0000 | 1.0000 | 1.0000 |
| 2 | 0.1417 | 0.2433 | 0.3682 | 0.4653 | 0.4976 | 0.5000 | 0.5000 | 0.5000 | 0.5000 |
| 4 | 0.1216 | 0.1841 | 0.2326 | 0.2488 | 0.2500 | 0.2500 | 0.2500 | 0.2500 | 0.2500 |
| 8 | 0.0921 | 0.1163 | 0.1244 | 0.1250 | 0.1250 | 0.1250 | 0.1250 | 0.1250 | 0.1250 |
| 16 | 0.0582 | 0.0622 | 0.0625 | 0.0625 | 0.0625 | 0.0625 | 0.0625 | 0.0625 | 0.0625 |
| 32 | 0.0311 | 0.0312 | 0.0312 | 0.0313 | 0.0313 | 0.0313 | 0.0313 | 0.0313 | 0.0313 |
| 64 | 0.0156 | 0.0156 | 0.0156 | 0.0156 | 0.0156 | 0.0156 | 0.0156 | 0.0156 | 0.0156 |
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