Wastewater Calculations - Mathematics and Statistics for the Environment - page 694

Solution:

BOD in = 220 mg/L × 1.1 MGD × 8.34 lb/gal = 2018 lb/day

BOD out = 18 mg/L × 1.1 MGD × 8.34 lb/gal = 165 lb/day

BOD removed = 2018 lb/day - 165 lb/day = 1853 lb/day

Solids produced = 1853 lb/day × 0.65 lb/lb BOD = 1204 lb/day

Solids out (lb/day) = 22 mg/L × 1.1 MGD × 8.34 lb/gal = 202 lb/day

Sludge out (lb/day) = 8710 mg/L × 0.024 MGD × 8.34 lb/gal = 1743 lb/day

Solids removed (lb/day) = 202 lb/day + 1743 lb/day =1945 lb/day

1204 lb/day − 1945 lb/day) × 100

%Massbalance = = 62%

1204 lb/day

The mass balance indicates the following:

1. The sampling points, collection methods, or laboratory testing procedures 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:

Solidsproduced (lb/day)
Waste T

Waste ( gpd ) = (24.50)

SS (mg/L) × 8.34

Thus,

1204 lb/day 1,000,000
8710 mg/L

×

Waste (gpd) = = 16,575 gpd

× 8.34

24.7.1.11 Aeration Tank Design Parameters
The two design parameters of aeration tanks are food-to-microorganism (F/M) ratio and aeration
period (similar to detention time). F/M ratio (BOD loading) is expressed as pounds of BOD per day
per pound of MLSS:

133,690(BOD)
(MLSS)

Q

o

F/Mratio = (2 4.51)

V

where
133,690 = Factor to convert units.
BOD = Settled BOD from primary tank (mg/L).
Q o = Average daily wastewater flow (MGD).
MLSS = Mixed liquor suspended solids (mg/L).
V = Volume of tank (ft 3 ).