(1258 Ib VSS/d)(2.3 Ib O 2 Ab VSS) = 2893 Ib O 2 /d (1313.4 kg/d)The volume of air required at standard conditions (14.7 lb/in^2 and 680 F) (96.5 kPa and
2O^0 C) assuming air contains 23.2 percent oxygen by weight and the density of air is 0.075
Ib/ft^3 is:
V°lume°f Ak - (0.075 (^8) Ib^
32 ) -
(^166) '264ft3/d(47° 53 m3/d)
For summer conditions:
- Oxygen required = (1447 lb/d)(2.3 Ib O 2 /d) = 3328 Ib O 2 /d (1510.9 kg/d)
- Volume of Air = 3328 Ib O 2 /d/(0.075 Ib/ft^3 )(0.232) = 191,264 ft^3 /d (5412.8 m^3 /d)
Note that the oxygen transfer efficiency of the digester system must be taken into ac-
count to get the actual volume of air required. Assuming diffused aeration with an oxygen
transfer efficiency of 10 percent, the actual air requirements at standard conditions are:
- Winter: volume of air = 166,264 ft^3 /d/(0.1)(1,400 min/d) = 1155 ftVmin (32.7 nWmin)
- Summer: volume of air = 191,264 ft^3 /d/(0.1)(1,440 min/d) = 1328 fVVmin (37.6
m^3 /min)
To summarize winter and summer conditions:
Parameter Winter Summer
Total Solids In, Ib/d (kg/d) 3933 (1785.6) 3933 (1785.6)
VSS In, Ib/d (kg/d) 3146 (1428.3) 3146 (1428.3)
VSS Reduction, (%) 40 46
VSS Reduction, Ib/d (kg/d) 1258 (571.1) 1447 (656.9)
Digested Sludge Out, gal/day (L/d) 6223 (23.6) 5790 (21.9)
Digested Sludge Out, Ib/d (kg/d) 2675(1214.5) 2486(1128.6)
Air Requirements @ S.C., fWmin (mVmin) 1155(32.7) 1328(37.6)
- Determine the aerobic digester volume
From the above analysis it is clear that the aerobic digester volume will be calculated us-
ing values obtained under the winter conditions analysis, while the aeration equipment
will be sized using the 1328 ft^3 /min (37.6 m^3 /min) air requirement obtained under the
summer conditions analysis.
The volume of the aerobic digester is computed using the following equation, assum-
ing the digester is loaded with waste activated sludge only:
y Q&
x(Kdpv+\iec)where V = Volume of aerobic digester, ft^3 (m^3 )
Qi = Influent average flow rate to the digester, fWd (m^3 /d)
X 1 = Influent suspended solids, mg/L (50,000 mg/L for 5.0% solids)