Environmental Engineering Reference
In-Depth Information
Solution:
The primary effluent and recirculated trickling filter effluent
are applied together across the surface of the filter; therefore,
0.488 MGD0.566 MGD1.054 MGD1,054,000 gpd
+
=
=
2
2
2
Circular surface area
=
0.785
×
(Diameter)
= 785 90ft)
0.
×
=
6359 ft
1,054,000 gpd
6359 ft
=165.7 gpd/ft
2
2
7.3.8.3 Organic Loading Rate
As mentioned earlier, trickling filters are sometimes classified by
the organic loading rate applied. The organic loading rate is expressed
as a certain amount of BOD applied to a certain volume of media.
Example 7.5
Problem:
A trickling filter, 50 ft in diameter, receives a primary effluent
flow rate of 0.445 MGD. Calculate the organic loading rate in units of
pounds of BOD applied per day per 1000 ft
3
of media volume. The pri-
mary effluent BOD concentration is 85 mg/L. The media depth is 9 ft.
Solution:
0.445 MGD × 85 mg/L × 8.34 lb/gal = 315.5 lb BOD/day
Surface Area = 0.785 × (50)
2
= 1962.5 ft
2
Area × Depth = Volume
1962.5 ft
2
× 9 ft = 17,662.5 ft
3
To determine the pounds of BOD per 1000 ft
3
in a volume of thousands
of cubic feet, we must set up the equation as shown below:
315.5lbBOD/day
17,662.5
1000
1000
×
Regrouping the numbers and the units together:
315.5lb
17,662.5ft
×
1000
lb BOD/day
1000 ft
3
×
=
17.9lbBOD/day/1000 ft
3
3
7.3.8.4 Settling Tanks
In the operation of settling tanks that follow trickling filters, vari-
ous calculations are routinely made to determine detention time, sur-
face settling rate, hydraulic loading, and sludge pumping.
