Geoscience Reference
In-Depth Information
EXAMPLE 24.23
Problem: A trickling filter 80 ft in diameter treats a primary effluent flow of 750,000 gpd. If the
recirculated flow to the clarifier is 0.2 MGD, what is the hydraulic loading on the trickling filter?
Solution:
750,000 gpd
0.785
149 gpd/ft 2
Hydraulic loadingrate
=
=
×
80
ft
×
80 ft
EXAMPLE 24.24
Problem: A high-rate trickling filter receives a daily flow of 1.8 MGD. What is the dynamic loading
rate in MGD/ac if the filter is 90 ft in diameter and 5 ft deep?
Solution:
2
0.785
×
90 ft
×
90 ft
=
6359 ft
2
6359 ft
43,560 ft
=
0.146 ac
2 /ac
1.8 MG
D
0.146 ac
Hydraulic loadingrate
=
=
12.3 MGD/ac
Key Point: When hydraulic loading rate is expressed as MGD per acre, this is still an expression of
gallon flow over surface area of the trickling filter.
24.5.1.2 Organic Loading Rate
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. In other words,
the organic loading is defined as the pounds of BOD 5 or chemical oxygen demand (COD) applied
per day per 1000 ft 3 of media—a measure of the amount of food being applied to the filter slime. To
calculate the organic loading on the trickling filter, two things must be known: the pounds of BOD
or COD being applied to the filter media per day and the volume of the filter media in 1000-ft 3 units.
The BOD and COD contribution of the recirculated flow is not included in the organic loading.
EXAMPLE 24.25
Problem: A trickling filter, 60 ft in diameter, receives a primary effluent flow rate of 0.440 MGD.
Calculate the organic loading rate in units of pounds of BOD applied per day per 1000 ft 3 of media
volume. The primary effluent BOD concentration is 80 mg/L. The media depth is 9 ft.
Solution:
0.440 MGD80mg/L
×
×
8.34 lb/gal
=
293.6 lb BOD
2
2
S
urface area = 0.785
×
(60)
=
2826 ft
2
3
Area
×
dep
th
=
Volume
=
2826 ft
×
9ft=25,434 ft
Note: 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 follows:
 
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