Primary Treatment (Sedimentation) - Wastewater Operators - page 254

 Example 6.4

Problem: The sludge pump operates 20 minutes per hour. The pump
delivers 20 gpm of sludge. Laboratory tests indicate that the sludge is
5.2% solids and 66% volatile matter. How many pounds of volatile matter
are transferred from the settling tank to the digester?

Solution:

Pump time = 20 min/hr
Pump rate = 20 gpm
Percent solids = 5.2%
Percent volatile matter = 66%

Volatile Solids = 20 gpm 20 min/hr × × 24 hr/day ))8.34lb/gal × × 0.052 × 0.66

= 2748 lb/day

6.3.6 Percent Total solids (%Ts)

 Example 6.5

Problem: A settling tank sludge sample is tested for solids. The sample
and dish weighed 74.69 g. The dish alone weighs 21.2 g. After drying, the
dish with the dry solids weighed 22.3 g. What is the percent total solids
(%TS) of the sample?

Solution:

Sample + dish 74.69 g Dish + dry solids 22.3 g

Dish alone -21.2 g Dish alone -21.2 g

Sample weight 53.49 g Dry solids weight 1.1 g

1.1g
53.49 g

× 100% = 2%

6.3.7 bod and ss removal (lb/day)

To calculate the pounds of biochemical oxygen demand (BOD) or sus-
pended solids (SS) removed each day, we need to know the mg/L BOD or
SS removed and the plant flow. Then, we can use the following equation:

SS Removed (lb/day) = SS (mg/L) × MGD × 8.34 lb/gal (6.5)

 Example 6.6

Problem: If 120 mg/L suspended solids are removed by a primary clari-
fier, how many lb/day suspended solids are removed when the flow is
6,250,000 gpd?

Solution:

SS Removed = 120 mg/L × 6.25 MGD × 8.34 lb/gal = 6255 lb/day