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4.2 Rectilinear Particle Motion 99

If the mouth is about 1.5 m above the floor, how long would it take for the 5-μm
droplet to settle down to the floor.

3 −5
Solution Given d p =5 μm, ρ p = 1,000 kg/m ,and μ = 1.81 × 10 Pa s,we can get,
0:066 lm
Kn ¼ 2k=d p ¼ 2 ¼ 0:0264
5 lm
0:999

C c ¼ 1 þ Kn 1:142 þ 0:558 exp ¼ 1:033
Kn
! 2 !
2 6
q d C c 1000 5 10 1:0333
p p
t ¼ 3s ¼ 3 ¼ 3 ¼ 0:00024 s
18l 18 1:81 10 5
2 6 2
q d gC c 1000 5 10 9:81 1:033
p
v TS ¼ ¼ 5 ¼ 0:000753 m=s
18l 18 1:81 10
H 1:5m
Dt ¼ ¼ ¼ 1992 s ¼ 33 min
v TS 0:000753 m/s
It could be much slower for a smaller particle.

4.2.2 Settling at High Reynolds Numbers

The particle motion is in Newton’s regime for Re [ 1:0 and the corresponding
settling speed in a gravitational field can be determined by equating the drag force
and the force of gravity, which gives

4q d g
p p
v TS ¼ ð4:24Þ
3C D q
g
where the drag coefficient C D was introduced before. However, it is not straight-
forward to find v TS using this equation. Because C D depends on the particle Rey-
nolds number. It cannot be determined without knowing v TS . One way to solve this
problem is an iterative solution obtained by substituting an initial guess of v TS into
the above equation and by trying different values of v TS until the solution converges
with a desired accuracy.