Environmental Engineering Reference
In-Depth Information
8.14.5.1.1 required mlvSS quantity
The pounds of MLVSS required in the aeration tank to achieve the
optimum F/M ratio can be determined from the average influent food
(BOD or COD) and the desired F/M ratio:
PrimaryEffluent BODorCOD
×
Flow
(MGD)
×
8.34
MLVSS (lb)
=
(8.12)
DesiredF/M Ratio
The required pounds of MLVSS determined by this calculation can then
be converted to a concentration value by:
DesiredMLVSS (lb)
Aeration Vol
MLVSS (mg/L)
=
(8.13)
ume(MG)8.34
×
Example 8.7
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/lb MLVSS. (1) How many pounds of MLVSS must be
maintained in the aeration tank to achieve the desired F/M ratio? (2)
What is the required concentration of MLVSS in the aeration tank?
Solution:
145mg/L4.0 MGD8.34lb/gal
0.3
× ×
lb COD/lbMLVSS
MLVSS (lb)
=
=
16,124 lb
16,1
24 MLVSS
0.65 MG
MLVSS (mg/L)
=
=
2974mg/L
×
8.34
8.14.5.1.2 Calculating Waste rates using the f/m ratio
Maintaining the desired F/M ratio is achieved by controlling the
MLVSS level in the aeration tank. This may be accomplished by adjust-
ment of return rates; however, the most practical method is by proper
control of the waste rate.
(8.14)
WasteVolatile Solids(lb/day)
=
Actual MLVSS
(lb)
DesiredMLVSS (lb)
If the desired MLVSS is greater than the actual MLVSS, wasting is
stopped until the desired level is achieved.
Practical considerations require that the required waste quantity
be converted to a required volume to waste per day. This is accomplished
by converting the waste pounds to flow rate in million gallons per day
(MGD) or gallons per minute (gpm):
WasteVolatileSolids(lb/day)
Waste(MGD)
=
(8.15)
W
asteVolatile Concentration(mg/L)8.34
×
 
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