Water Treatment Processes - Mathematics and Statistics for the Environment - page 652

23.7.1 C hlorine d isinFeCtion
Chlorine can destroy most biological contaminants by various mechanisms, including

• Damaging the cell wall
• Altering the permeability of the cell (the ability to pass water in and out through the cell
wall)
• Altering the cell protoplasm
• Inhibiting the enzyme activity of the cell so it is unable to use its food to produce energy
• Inhibiting cell reproduction

Chlorine is available in a number of different forms: (1) as pure elemental gaseous chlorine (a green-
ish-yellow gas possessing a pungent and irritating odor that is heavier than air, nonflammable, and
nonexplosive), which, when released to the atmosphere, is toxic and corrosive; (2) as solid calcium
hypochlorite (in tablets or granules); or (3) as a liquid sodium hypochlorite solution (in various
strengths). The choice of one form of chlorine over another for a given water system depends on
the amount of water to be treated, configuration of the water system, the local availability of the
chemicals, and the skill of the operator. One of the major advantages of using chlorine is the effective
residual that it produces. A residual indicates that disinfection is completed and the system has an
acceptable bacteriological quality. Maintaining a residual in the distribution system helps to prevent
regrowth of those microorganisms that were injured but not killed during the initial disinfection stage.

23.7.2 d etermining C hlorine F eed r ate
The expressions milligrams per liter (mg/L) and pounds per day (lb/day) are most often used to
describe the amount of chlorine added or required. Equation 23.87 can be used to calculate either
mg/L or lb/day chlorine dosage:

Chlorine feed rate (lb/day) = Chlorine (mg/L) × Flow (MGD) × 8.34 lb/gal (23.87)

■ EXAMPLE 23.103
Problem: Determine the chlorinator setting (lb/day) required to treat a flow of 4 MGD with a chlo-
rine dose of 5 mg/L.

Solution:

Chlorine feed rate = Chlorine (mg/L) × Flow (MGD) × 8.34 lb/gal

= 5 mg/L × 4 MGD × 8.34 lb/gal = 167 lb/day

■ EXAMPLE 23.104
Problem: A pipeline that is 12 in. in diameter and 1400 ft long is to be treated with a chlorine dose
of 48 mg/L. How many pounds of chlorine will this require?

Solution: First determine the gallon volume of the pipeline:

Volume = 0.785 × (Diameter) 2 × Length (ft) × 7.48 ga l /f (ft) 3

= 0.785 × (1 ft) 2 × 1400 ft × 7.48 gal l /f ft 3 = 8221 gal

Now calculate the pounds chlorine required:

Chlorine = Chlorine (mg/L) × Volume (MG) × 8.34 lb/gal

= 48 mg/L × 0.008221 MG × 8.34 lb/gal = 3.3 lb