Fig. 157.—Diagram for the Computation of Economical Basin Dimensions.
For example, let it be desired to determine the dimensions of two continuous-flow sedimentation basins as shown in Fig. 157, in which the period of retention in each is to be 2 hours, the velocity of flow is not to exceed one foot per second, and the sludge accumulated will be 3 cubic yards per million gallons of sewage treated. The quantity of sewage to be treated is 18,000,000 gallons per day. The shortest time between cleanings will be 2 weeks.
The capacity of each basin must be 2
24 of 18,000,000 gallons, or 200,000 cubic feet in order to allow a period of retention of 2 hours. To this volume should be added sufficient capacity to allow for the 2 weeks of sludge storage between cleanings. When a basin is being cleaned the load must be put on the remaining basins. Then if Q represents the rate of accumulation of sludge per day, n represents the number of days between cleanings, m represents the number of basins, and S the sludge capacity of one basin, then
S = Q(n − 1)
m + Q
m − 1
The sludge storage capacity for the example given will be approximately 11,000 cubic feet.
In expressing the total cost of the basins let
h = the depth in feet.
l = the length in feet.
b = the width in feet.
| The cost of land, floor, etc., per square foot | = p dollars. |
| The cost of wall per foot length | = qh2 dollars. |
| The cost of pipes, valves and appurtenances | = P dollars. |
| Then the total cost C = (3l + 4b)qh2 + 2plb + P. | |