Aerosol Kinetics
Terminal Velocity
For a particle starting from rest, the solution to equiaton 2 is given as
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(5)
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where uƒ is assumed to be a constant
vector. For uƒ=0 and large t, the terminal velocity
of particle ut is given by
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(6)
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Table 7: Relaxtion time τ for a unit density particle in the air(p=1 atm, T=20°C)
Diameter, µm | ut=τ g | τ (sec) | Stop Distance uo = 1m/s |
Stop Distance uo=10m/s |
0.05 | 0.39 µm/s | 4x10-8 | 0.04 µm | 4x10-4 mm |
0.1 | 0.93 µm/s | 9.15-8 | 0.092 µm | 9.15x10-4 mm |
0.5 | 10.1 µm/s | 1.03x10-6 | 1.03 µm | 0.0103 mm |
1 | 35 µm/s | 3.57x10-6 | 3.6 4µm | 0.0357 mm |
5 | 0.77 mm/s | 7.86x10-5 | 78.6 µm | 0.786 mm |
10 | 3.03 mm/s | 3.09x10-4 | 309 µm | 3.09 mm |
50 | 7.47 cm/s | 7.62x10-3 | 7.62 mm | 76.2 mm |
Stopping Distance
In the absence of gravity and fluid flow, for a particle with an initial veolocity of u,
the solution to 2 is given by
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(7)
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(8)
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where xp is the position of the particle.
As t→&infin, up→0 and
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(9)
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is known as the stopping distance of the particle. For an initial
veolocity of 1000 cm/s, the stop distance for various particles is
listed in Table 7.
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