Geoscience Reference
In-Depth Information
24.7.1.4.1 Required MLVSS Quantity
The pounds of MLVSS required in the aeration tank to achieve the optimum F/M ratio can be deter-
mined from the average influent food (BOD or COD) and the desired F/M ratio:
Primaryeffluent BOD/COD low(M
×
GD)
×
8.34 lb/gal
MLVSS(lb)
=
(24.35)
DesiredF/M ratio
The required pounds of MLVSS determined by this calculation can then be converted to a concen-
tration value by
DesiredMLVSS (lb)
Aeration vol
MLVSS(mg/L)
=
(24.36)
ume(MG)
×
8.34 lb/gal
EXAMPLE 24.45
Problem: The aeration tank influent flow is 4.0 MGD and the influent COD is 145 mg/L. The
aeration tank volume is 0.65 MG. The desired F/M ratio is 0.3 lb COD per lb MLVSS. How many
pounds of MLVSS must be maintained in the aeration tank to achieve the desired F/M ratio?
Solution: Determine the required concentration of MLVSS in the aeration tank:
145 mg/L
× ×
lb CODper lb MLVSS
4.0 MGD
8.34 lb/gal
MLVSS(lb)
=
=
16,124 lb
0.3
16,124 lb
0.65 MG
MLVSS(mg/L)
=
= 974 mg/L
2
×
8.34 lb/gal
24.7.1.4.2 Calculating Waste Rates Using F/M Ratio
Maintaining the desired F/M ratio is accomplished by controlling the MLVSS level in the aeration
tank. This may be achieved by adjustment of return rates; however, the most practical method is by
proper control of the waste rate:
Waste volume solids (lb/day) = Actual MLVSS (lb/day) - Desired MLVSS (lb/day)
(24.37)
If the desired MLVSS is greater than the actual MLVSS, wasting is stopped until the desired level is
achieved. Practical considerations require that the waste quantity be converted to a required volume
of waste per day. This is accomplished by converting the waste pounds to flow rate in million gal-
lons per day or gallons per minute:
Waste volatile(lb/day)
Waste (MGD)
=
(24.38)
Waste vo
latile concentration(mg/L)
×
8.34 lb/gal
Waste (MGD) 1,000,000 gpd/MGD
1440
×
Waste (gpm)
=
(24.39)
min/day
Key Point: When the F/M ratio is used for process control, the volatile content of the waste activated
sludge should be determined.
 
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