Water Treatment Processes - Mathematics and Statistics for the Environment - page 631

■ EXAMPLE 23.62
Problem: The flow to a solids contact clarifier is 2,650,000 gpd. If the lime dose required is deter-
mined to be 12.6 mg/L, how many lb/day lime will be required?

Solution:

Lime (lb/day) = Lime (mg/L) × flow (MGD) × 8.34 lb/gal

= 12.6 mg/L × 2.65 MGD × 8.34 lb/gal = 278 lb/day lime

23.5.9 d etermining l ime d osage (g rams per m inute )
To convert from mg/L lime to g/min lime, use Equation 23.58:

Key Point: 1 lb = 453.6 g.

Lime (lb/day) ×

in/day

453.6 g/lb

Lime (g/min) = (23.58)

1440 m

■ EXAMPLE 23.63
Problem: A total of 275 lb/day lime will be required to raise the alkalinity of the water passing
through a solids-contact clarification process. How many g/min lime does this represent?

Solution:

Lime (lb/day) 453.6 g/lb
1440 min/day

× 2 75 lb/day 453.6 g/lb

1440 min/day

×

Lime = = = 86.6 g/mi n

■ EXAMPLE 23.64
Problem: A lime dose of 150 lb/day is required for a solids-contact clarification process. How many
g/min lime does this represent?

Solution:

Lime (lb/day) 453.6 g/lb
1440 min/day

× 1 50 lb/day 453.6 g/lb

1440 min/day

×

Lime = = = 47.3 g/mi n

23.5.10 p artiCle s ettling (s edimentation ) *
Particle settling (sedimentation) may be described for a singular particle by Newton's equation
(Equation 23.64) for terminal settling velocity of a spherical particle. For the engineer, knowledge
of this velocity is basic in the design and performance of a sedimentation basin. The rate at which
discrete particles will settle in a fluid of constant temperature is given by the following equation:

( )

1/2

4 gp pd

Cp

−

p

u = (23.59)

3

D

* Much of the information presented in this section is based on USEPA, Guidance Manual for Compliance with the Interim
Enhanced Surface Water Treatment Rule: Turbidity Provisions , EPA 815-R-99-012, U.S. Environmental Protection
Agency, Washington, DC 1999.