Wastewater Calculations - Mathematics and Statistics for the Environment - page 715

24.11 BIOSOLIDS PRODUCTION AND PUMPING

24.11.1 p roCess r esiduals
The wastewater unit treatment processes remove solids and biochemical oxygen demand from the
wastestream before the liquid effluent is discharged to its receiving waters. What remains to be
disposed of is a mixture of solids and wastes, called process residuals , more commonly referred to
as biosolids (or sludge ).

Note: Sludge is the commonly accepted term for wastewater residual solids; however, if wastewater
sludge is used for beneficial reuse (i.e., as a soil amendment or fertilizer), it is commonly
called biosolids . I choose to refer to process residuals as biosolids in this text.

The most costly and complex aspect of wastewater treatment can be the collection, processing,
and disposal of biosolids. The quantity of biosolids produced may be as high as 2% of the origi-
nal volume of wastewater, depending somewhat on the treatment process being used. Because the
biosolids can be as much as 97% water content and because cost of disposal will be related to the
volume of biosolids being processed, one of the primary purposes or goals (along with stabilizing it
so it is no longer objectionable or environmentally damaging) of biosolids treatment is to separate
as much of the water from the solids as possible.

24.11.2 p rimary and s eCondary s olids p roduCtion C alCulations
It is important to point out that when making calculations pertaining to solids and biosolids, the
term solids refers to dry solids and the term biosolids refers to the solids and water. The solids pro-
duced during primary treatment depend on the solids that settle in, or are removed by, the primary
clarifier. When making primary clarifier solids production calculations, we use the mg/L to lb/day
equation for suspended solids as shown below:

SS removed (lb/day) = SS removed (mg/L) × Flow (MGD) × 8.34 lb/gal (24.113)

24.11.3 p rimary C lariFier s olids p roduCtion C alCulations

■ EXAMPLE 24.82
Problem: A primary clarifier receives a flow of 1.80 MGD with suspended solids concentrations
of 340 mg/L. If the clarifier effluent has a suspended solids concentration of 180 mg/L, how many
pounds of solids are generated daily?

Solution:

SS removed (lb/day) = SS removed (mg/L) × Flow (MGD) × 8.34 lb/gal

= 160 mg/L × 1.80 MGD × 8.34 lb/gal

= 2402 lb/day

■ EXAMPLE 24.83
Problem: The suspended solids content of the primary influent is 350 mg/L and the primary influ-
ent is 202 mg/L. How many pounds of solids are produced during a day when the flow is 4,150,000
gpd?