Page 149 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

104 Fundamentals of Water Treatment Unit Processes: Physical, Chemical, and Biological

1=24
2=20
v 2 3=24
v H 4=18
D 5=20
v 3 v R2
Total = 106
d 3
v H
v R3

r = d /D ·ΔP L
3
3
3
= 3/5·24 particles
= 15 particles

FIGURE 6.11 Ideal settling basin showing settling paths of the particles of the ‘‘3’’ size fraction. All size ‘‘3’’ particles are removed below
level d 3 . Overflow velocity, v o , equals fall velocity for size ‘‘2’’ particles, i.e., v 2 .

can see that a size 2 particle located at the top of the entrance these particles, i.e., v s < v o , consider, as in Figure 6.10b, a
cross section will just reach the bottom of the basin after particle size designated ‘‘1,’’ which has an associated size
distance, L.Bydefinition then, v 2 ¼ v o , i.e., the ‘‘overflow fraction, DP 1 . The proportion of those particles removed,
=
velocity.’’ r 1 ¼ (v 1 v o ) DP 1 , as seen in Figure 6.10b. Figure 6.11 illus-
The main idea of Figure 6.11 is to provide a means to trates the concept intuitively. The removal of all particles
better visualize the idea of a specific particle fraction, e.g., where v s < v o is the sum of the infinitesimal removals, r i ,
Ð
= P o v s dP, which is the shaded area shown in Figure
DP i , and the proportion removed, e.g., r 3 ¼ (d 3 D) DP 3 i.e., 1=v o 0
(Example 6.3). 6.10b divided by v o . The total removal, R, is the sum of the
integral and the quantity, (1 P o ), and is stated as
Example 6.3 Illustrate the Fraction Removals P ð o
for a Heterogeneous Suspension v s dP (6:18)
R ¼ (1 P o ) þ
v o
0
Given
Suppose there are 24 particles having fall velocity, v 3 , i.e., where
DP 3 ¼ 24 particles (in Figure 6.10, this is the count of size
3 particles). Also suppose that d 3 =D ¼ 3=5 (measuring R is the total fraction of suspension removed in the ideal
these distances in Figure 6.10). settling basin
v o is the overflow velocity (m=s)
Solution v s is the fall velocity of any particle in suspension (m=s)
Based upon the mathematics noted previously, the number
P is the fraction of suspension associated with any v s
of d 3 particles removed is r 3 ¼ (d 3 =D) DP 3 ¼ 3=5 24 ¼ 15.
P o is the fraction of suspension associated with overflow
Discussion velocity, v o
Counting the size ‘‘3’’ particles confirms that there are
15 particles coming into the basin that are at depth d 3 . To apply Equation 6.18, the integral must be evaluated
Also, to reinforce the idea that fall velocities may be used
graphically. The problems are for two cases: (1) R is specified
for the same calculation, since d 3 =D ¼ v 3 =v o , and since fall and v o must be determined, and (2) v o is specified and R must
velocities are used to characterize a suspension, we can
also use the expression, r 3 ¼ (v 3 =v o ) DP 3 in lieu of d 3 =D. be determined. For Case 1, different v o values are assumed
and trial-and-error calculations are done until the right-hand
side of Equation 6.18 satisfies the specified R. For Case 2, the
determination is a straightforward graphical integration and
6.4.3 MATHEMATICS OF REMOVAL
the value of R is whatever is calculated. Figure 6.12 shows
Figure 6.10b indicates the method of calculation of the total results of a settling column analysis of a hypothetical discrete
removal, R. For a given overflow velocity, v o , all particles particle suspension (Camp, 1946), as illustrated in Figure 6.10b.
where v s v o , or the (1 P o ) fraction, will be 100% removed. For any assumed overflow velocity, e.g., v o ¼ 0.11 cm=s, with an
Particles with v s < v o will be partially removed. Regarding associated P o ¼ 0.94, all particles with v s 0.11 cm=s are