Environmental Engineering Reference
In-Depth Information
Example 8.4
Problem: The operator wastes 0.44 MGD of activated sludge. The waste
activated sludge has a solids concentration of 5540 mg/L. How many
pounds of waste activated sludge are removed from the process?
Solution:
Waste = 5540 mg/L × 0.44 MGD × 8.34 lb/MG/mg/L = 20,329.6 lb/day
8.14.5 food-to-Microorganisms ratio
The food-to-microorganisms (F/M) ratio is a process control calcu-
lation used in many activated sludge facilities to control the balance
between available food materials (BOD or COD) and available organisms
(mixed liquor volatile suspended solids). The chemical oxygen demand
(COD) test is sometimes used, because the results are available in a
relatively short period of time. To calculate the F/M ratio, the following
information is required:
Aeration tank
influent flow rate (MGD)
Aeration tank influent BOD or
COD (mg/L)
Aeration tank MLVSS (mg/L)
Aeration tank volume (MG)
Primary Effluent COD/BOD (mg/L)
F
) Aerator Volume (MG)
8.34 lb/MG/mg/L
× low (MGD)
×
8.34 lb/MG/mg/L
F/M Ratio
=
(8.10)
MLVSS (mg/L
×
⎥⎥
×
Typical F/M ratios for activated sludge processes are shown below:
Process
lb bod 5 /lb Mlvss
lb Cod/lb Mlvss
Conventional
0.2-0.4
0.5-1.0
Contact stabilization
0.2-0.6
0.5-1.0
Extended aeration
0.05-0.15
0.2-0.5
Oxidation ditch
0.05-0.15
0.2-0.5
Pure oxygen
0.25-1.0
0.5-2.0
Example 8.5
Problem: Given the following data, what is the F/M ratio?
Primary effluent flow = 2.5 MGD
Aeration volume = 0.65 MG
Primary effluent BOD = 145 mg/L
Settling volume = 0.30 MG
Primary effluent TSS = 165 mg/L
MLSS = 3650 mg/L
Effluent flow = 2.2 MGD
MLVSS = 2550 mg/L
Effluent BOD = 22 mg/L
% Waste VM = 71%
Effluent TSS = 16 mg/L
Desired F/M = 0.3
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