Wastewater Calculations - Mathematics and Statistics for the Environment - page 721

24.12.1.1 Estimating Daily Biosolids Production
The calculation for estimation of the required biosolids pumping rate provides a method to establish
an initial pumping rate or to evaluate the adequacy of the current pump rate:

(Influent TSSconc. − EluentTSS conc.) ff × Flow × 8.34

Estimated pumprate = (24.126)

%Solidsinbios olids ×× 8.34 1440 min/day

■ EXAMPLE 24.95
Problem: The biosolids withdrawn from the primary settling tank contain 1.5% solids. The unit
influent contains 280 mg/L TSS, and the effluent contains 141 mg/L TSS. If the influent flow rate
is 5.55 MGD, what is the estimated biosolids withdrawal rate in gallons per minute (assuming that
the pump operates continuously)?

Solution:

(280 mg/L − 141 mg/ L) × 5.55 MGD × 8.34

Biosolidswithdrawal rate = = 36 gpm

0.015 ×× 8.34 1440 min/day

24.12.1.2 Surface Loading Rate (gal/day/ft 2 )
The surface loading rate (surface settling rate) is hydraulic loading—the amount of biosolids applied
per square foot of gravity thickener:

Biosolidsapp lied (gpd)

Thickenerarea (ft )

2 =

Surface loading(gal/day/ft) (24.127)

2

■ EXAMPLE 24.96
Problem: A 70-ft-diameter gravity thickener receives 32,000 gpd of biosolids. What is the surface
loading in gallons per square foot per day?

Solution:

32,000 gpd
0.785

= 8.32 gpd/ft 2 Surface loading =

× 70 ft × 70 ft

24.12.1.3 Solids Loading Rate (lb/day/ft 2 )
The solids loading rate is the pounds of solids per day being applied to 1 ft 2 of tank surface area.
The calculation uses the surface area of the bottom of the tank. It assumes that the floor of the tank
is flat and has the same dimensions as the surface:

%Biosolid sssolids × B

hickener area (ft )

iosolidsflow(gpd) × 8.34 lb/gal

2 =

Solids loadingrate(lb/day/ft) (24.128)

2

T

■ EXAMPLE 24.97
Problem: The thickener influent contains 1.6% solids. The influent flow rate is 39,000 gpd. The
thickener is 50 ft in diameter and 10 ft deep. What is the solid loading in pounds per day?

Solution:

0.016 × 39,000 gpd × 8.34 l b/gal

2.7 lb/day/ft 2 Solids loadingrate = =

0.785 × 50 ft × 50 ft