TM 5-814-3/AFM 88-11, Volume III
(6) Calculate sludge volume, assuming a specific gravity of 1.02 and a moisture content of 96% (4%
Sludge volume '
' 2,360 cu ft/day (@4 mgd) ' 17,700gpd.
1.20 (62.4 lb/cu ft) (0.04)
(7) Sludge handling in this example consists of removing sludge manually from settling tank sludge
hopper, using a telescoping drawoff pipe which discharges the sludge into a sump from which it is removed
by a sludge pump (or pumps). Assume that the sludge will be wasted every 8 hours and pumps for -hour
to the digester.
daily sludge volume
2,360 cu ft
Sludge sump capacity '
' 787 cu ft (5,900 gal).
Number of wasting periods per day
Increase capacity 10 percent to compensate for scum removal volumes:
Sludgeand scum volume/wasting period
' 217; use 220 gpm.
Sludge pumping capacity '
30 minutes pumping/wasting period
a. Design requirements and criteria. Calculate the sludge production, using chemical addition in
primary sedimentation. Assume that addition of 60 lbs of ferrous sulfate and 700 lbs/mil gal of lime yields 70
percent suspended solids removal under the following conditions:
Flow rate = 4 mgd;
Suspended solids concentration = 300 mg/L.
The reactions that will occur are as follows:
FeSO4.7H20 + Ca(HO3) Fe(HCO3)2 + CaSO4 + 7H2O
If lime in the form of Ca(OH)2 is added, the following reaction occurs:
Fe(HCO3)2 + 2Ca(OH)2 + Fe(OH)2 + 2CaCo3 + 2H2O
The ferrous hydroxide is next oxidized to ferric hydroxide by the dissolved oxygen in the sewage:
Fe(OH)2 + O2 + 2H2O 4Fe(OH)3.
The reaction of quicklime with water alkalinity and carbon dioxide:
CaO + H2O CA(OH)2 hydrated lime;
Ca(OH)2 + CO2 CaCO3 + H2O;
Ca(OH)2 + Ca(HCO3)2 2CaCo3 + 2H2O.
b. Calculations and results. All interim calculations are computed on the basis of a flow volume of 1 mil
(1) Determine the weight of suspended solids removed:
8.34 lb/mil gal
Solids weight ' (0.70)(300 mg/L)
' 1,750 lb/mil gal;
(2) Determine weight of ferric hydroxide formed from ferrous sulfate: