U s ¼
2
9
q p
À Á
mq a
gr
2
ð6:138Þ
where the kinematic viscosity m of air is assumed as 16*10
–6 m
2 s
−1 , q a is the air
density assumed as 1.16 kgm
−3
, g the gravitational acceleration of 9.8 ms
−2 , the
particle radius is 5*10
–6 m, and the density of particle r p of 1.28 gcm
−3 is given by
1.28 * 10
–3 * 10
6 = 1.28 * 10Kgm
−3 .
Inserting these values in Eq. (6.138), we have
U s ¼
2
9
1:28 Ã 10
3
À
Á Ã 9:8 Ã ð5 Ã 10
À6
Þ
2
1:164Ãð16 Ã 10
À6
Þ
$ 0:003ms
À1
¼ 3mms
À1
The corresponding Reynolds number of particle Re p of 2rU/m is
Re p ¼
2 Ã 0:003 Ã 5 Ã 10
À6
ð16 Ã 10
À6
Þ
$ 0:002
A value within the Stokes regime (Sect. 6.5.1) justifying the application of
Eq. (6.138).
(2) Applying Stokes formulation, Eq. (6.138), for the estimation of sedimentation
velocity, we have
U s ¼
2
9
910 Ã 9:8 Ã ð0:005Þ
2
1:16 Ã ð16 Ã 10
À6
Þ
$ 2669 ms
À1
corresponding to a Reynolds number of particles of
Re p ¼
2 Ã 0:005 Ã 2669
ð16 Ã 10
À6
Þ
$ 26:7
This value fails the transitional regime between Stokes and inertial domain, so
that the Stokes formulation, Eq. (6.138), is not valid, for the estimation of the
sedimentation velocity requiring a trial-and-error iterative approach until achieving
almost equilibrium between drag and gravitational forces. The main equations now
relevant are the following:
F d ¼ 0:5c d q g V
2
s pR
2
ð6:132Þ
F g ¼
4
3
pR
3 gðq pÀ q f Þ
ð 6:135Þ
U
2
s ’ 8Rq P g= 3c d q f
À
Á
ð6:137Þ
7.13 Example 12: Calculation of Sedimentation Velocity of a Particle
263