Page 99 - Modelling in Transport Phenomena A Conceptual Approach

4.3. FLOW PAST A SINGLE SPHERE 79

Analysis

For the minimum particle size that can be removed with 100% eficiency, the time
required for this particle to fall a distance H must be equal to the time required to
moue this particle horizontally a distance L, i.e.,

where (v) represents the average gas velocity in the settling chamber. Taking
(v) = 3 m/ s, the settling velocity of the particles can be calculated as

= (3) (:) = 0.43m/s

The value of Y is calculated from Eq. (4.3-15) as

- - (9.8)(2200 - 1.1845)(18.41 x
4
- = 4.74
3 ( 1.1845)2(0.43)3
Substitution of the value of Y into Eq. (4.3-18) gives

0.00019
0.052 + 0.007 -)
-
3.15+ ~1/4 -
y1/2 y3/4
0.00019 1
= exp 3.15 + (4.74)1/4 + (4.74)’P - (4.74)3/4 = 24.3
0.052
[
0.0°7
Therefore, the Reynolds number and the particle diameter are
V)
Rep =
(6 y13/20 - YS/ll
) l7l2O
24.3
- 17/20 = 2*55
-
[6 (4.74)l3lzo - (4.74) ]
PbP
Dp = -
P vt
- (18.41 x 10-6)(2.55) = 92 x 10-6m
-
(1.1845) (0.43)