Environmental Engineering Reference
In-Depth Information
Effluent flow = 1.5 MGD
Effluent flow BOD
5
= 18 mg/L
Effluent flow TSS = 22 mg/L
Waste flow = 24,000 gpd
Waste TSS = 8710 mg/L
Solution:
BODIn 20 mg/L
=
×
1.1MGD
×
8.34
=
2018 lb/day
5
BO
DDOut
=
18 mg/L
×
1.1MGD
×
8.34
=
165lb/day
5
BOD
Removed
=
2018 lb/day
−
165 lb/day
=
1853lb/da
y
5
Solids Produced
=
1853 lb/day
×
0.65 lb/lbBO
D5
=
1204 lb/day
Solids Out 2mg/L1.1 MGD8
= × × .34 02 lb/day
Sludge Out 710mg/L0.024 M
=
=
×
GD
×
8.34
=
1743 lb/day
Solids Removed
=
(292 lb/
day
+
1743 lb/day)
=
1945 lb/day
=
1204 lb/day
(
−
1945 lb/day)
×
100
= %
62
Mass Balance
1204 lb/day
The mass balance indicates either of the following:
1. The sampling points, collection methods, or laboratory testing pro-
cedures are producing nonrepresentative results.
2. The process is removing significantly more solids than is required.
Additional testing should be performed to isolate the specific cause
of the imbalance.
To assist in the evaluation, the waste rate based on the mass balance
information can be calculated:
Solids Produced (lb/day)
WasteT
Waste(gpd)
=
(8.26)
SS (mg/L)
×
8.34
1204lb/day1,000,000
8710 mg/L
×
Waste
=
= 6,575gpd
1
×
8.34
8.14.8 solids Concentration in secondary Clarifier
The solids concentration in the secondary clarifier can be assumed
to be equal to the solids concentration in the aeration tank effluent. It
may also be determined in the laboratory using a core sample taken
from the secondary clarifier. The secondary clarifier solids concentra-
tion can be calculated as an average of the secondary effluent suspended
solids and the return activated sludge suspended solids concentration.
