Geoscience Reference
In-Depth Information
Chemical (lb) = Chemical (mg/L) × Tank volume (MG) × 8.34 lb/gal
(24.90)
EXAMPLE 24.62
Problem: To neutralize a sour digester, 1 lb of lime is added for every pound of volatile acids in the
digester biosolids. If the digester contains 300,000 gal of biosolids with a volatile acid level of 2200
mg/L, how many pounds of lime should be added?
Solution: Because the volatile acid concentration is 2200 mg/L, the lime concentration should also
be 2200 mg/L, so
Lime required = Lime concentration × Digester volume × 8.34 lb/gal
= 2200 mg/L × 0.30 MG × 8.34 lb/gal = 5504 lb
24.10.2 C hlorine d ose , d emand , and r esidual
Chlorine is a powerful oxidizer that is commonly used in wastewater and water treatment for disin-
fection and in wastewater treatment for odor and bulking control, among other applications. When
chlorine is added to a unit process, we want to ensure that a measured amount is added, obviously.
Chlorine dose depends on two considerations: the chlorine demand and the desired chlorine residual:
Chlorine dose = Chlorine demand + Chlorine residual
(24.91)
24.10.2.1 Chlorine Dose
To describe the amount of chemical added or required, we use Equation 24.92:
Chemical (lb/day) = Chemical (mg/L) × Flow (MGD) × 8.34 lb/day
(24.92)
EXAMPLE 24.63
Problem: Determine the chlorinator setting (lb/day) required to treat a flow of 8 MGD with a chlo-
rine dose of 6 mg/L.
Solution:
Chemical = Chemical × Flow × 8.34 lb/day = 6 mg/L × 8 MGD × 8.34 lb/gal = 400 lb/day
24.10.2.2 Chlorine Demand
The chlorine demand is the amount of chlorine used in reacting with various components of the
water such as harmful organisms and other organic and inorganic substances. When the chlorine
demand has been satisfied, these reactions cease.
EXAMPLE 24.64
Problem: The chlorine dosage for a secondary effluent is 6 mg/L. If the chlorine residual after a
30-min contact time is found to be 0.5 mg/L, what is the chlorine demand expressed in mg/L?
Solution:
Chlorine dose = Chlorine demand + chlorine residual
6 mg/L = x mg/L + 0.5 mg/L
6 mg/L - 0.5 mg/L = x mg/L
x = 5.5 mg/L chlorine demand
Search WWH ::




Custom Search