Page 269 - Water and wastewater engineering

6-46 WATER AND WASTEWATER ENGINEERING

Solution:
a. To provide redundancy, the flow will be divided into two flocculation basins. The flow
in each is
3 3
50 10 m /d 3 3
Q each 25 10 m /d
2

b. Using Equation 6-13, calculate the volume of a flocculator basin as
3
3
(25 10 m /d ) 3
V (22 min ) 381 94 o. r r 382 m
1 440 min/d,
c. With three compartments, each compartment volume is

382 m 3 3
V 127 33 m.
3
d. Assuming a trial water depth of 4.0 m, the surface area of a compartment is

3
127 33 m 2
.
.
A surface 3183 m
40 . m
e. With a basin depth of 4.0 m and the design criterion of a basin depth 1 m greater than the
diameter of the paddle wheel, the paddle wheel diameter is
.
.
40 m 1 m 3 0 m
f. The minimum length of a stage (compartment) is the diameter of the paddle wheel plus
the design criterion of 1 m minimum between stages.
.
.
Minimum length 3 0 m 1 m 4 0 m
g. The nominal width of a compartment is the surface area divided by the minimum length
of a stage

.
3183 m 2
W 796 m
.
40 . m
h. The required clearance is twice the minimum clearance from the walls plus 1 m between
paddle wheels on a shaft.
.
)
Required clearance 20 3 m 1 m 1 60 m
(
.

i. The total length of the two paddle wheels is then
.
.
.
796 m 1 60 m 6 36 m