Water Treatment Processes - Mathematics and Statistics for the Environment - page 614

■ EXAMPLE 23.22
Problem: Alum dosage is 40 mg/L and K = 90 per day based on lab tests. Compute the detention
times for complete mixing and plug flow reactor for 90% reduction.

Solution: First find C e :

C e = (1 - 0.9) × C i = 0.1 × C i = 0.1 × 40 mg/L = 4 mg/L

Now calculate t for complete mixing (Equation 23.20):

V
QK

1 CC

C

−

=

1
90/day

40 mg/L 4mg/L
4mg/L

− 1

90

d 1440 mi n

= 144 min

==

=

×

i e

t

1day

e

Finally, calculate t for plug flow using the following formula:

1 C

C

1440
90

40
4

=

=

= 36.8 min

i
e

t ln ln

K

23.4.3 F loCCulation
Flocculation follows coagulation in the conventional water treatment process. Flocculation is the
physical process of slowly mixing the coagulated water to increase the probability of particle col-
lision. Through experience, we see that effective mixing reduces the required amount of chemi-
cals and greatly improves the sedimentation process, which results in longer filter runs and higher
quality finished water. The goal of flocculation is to form a uniform, feather-like material similar
to snowflakes—a dense, tenacious floc that traps the fine, suspended, and colloidal particles and
carries them down rapidly into the settling basin. To increase the speed of floc formation and the
strength and weight of the floc, polymers are often added.

23.4.4 C oagulation and F loCCulation g eneral C alCulations
Proper operation of the coagulation and flocculation unit processes requires calculations to deter-
mine chamber or basin volume, chemical feed calibration, chemical feeder settings, and detention
time.

23.4.4.1 Chamber and Basin Volume Calculations
To determine the volume of a square or rectangular chamber or basin, we use Equation 23.22 or
Equation 23.23:

Volume (ft 3 ) = Length (ft) × Width (ft) × Depth (ft) (23.22)

Volume (gal) = Length (ft) × Width (ft) × Depth (ft) × 7.48 (gal) l /f (ft) 3 (23.23)

■ EXAMPLE 23.23
Problem: A flash mix chamber is 4 ft square with water to a depth of 3 ft. What is the volume of
water (in gallons) in the chamber?

Solution:

Volume = Length (ft) × Width (ft) × Depth (ft) × 7.48 ga l /f (ft) 3

= 4 ft × 4 ft × 3 ft × 7.48 gal l /f ft 3 = 359 gal