Wastewater Calculations - Mathematics and Statistics for the Environment - page 710

24.10.7 C hemiCal s olution F eeder s etting ( gpd )
Calculating a gallon-per-day feeder setting depends on how the solution concentration is expressed:
lb/gal or percent. If the solution strength is expressed as lb/gal, use the following equation:

Chemical (mg/L) × × .34 lb/gal

Chemical solution(lb)

flow (MGD)8

Solution (gpd) = (24.100)

In water and wastewater operations, a standard, trial-and-error method known as jar testing is
conducted to determine optimum chemical dosage. Jar testing has been the accepted bench testing
procedure for many years. After jar testing results are analyzed to determine the best chemical dos-
age, the actual calculations are made, as demonstrated by the following example problems.

■ EXAMPLE 24.72
Problem: Jar tests indicate that the best liquid alum dose for a water is 8 mg/L. The flow to be
treated is 1.85 MGD. Determine the gpd setting for the liquid alum chemical feeder if the liquid
alum contains 5.30 lb of alum per gallon of solution.

Solution: First, calculate the lb/day of dry alum required, using the mg/L to lb/day equation:

Dose (mg/L) × flow (MGD) ×

ical solution(lb)

8.34 lb/gal

Alum =

Chem

= 8mg/L × 1.85 MGD × 8.34 lb /gal = 123 lb/day

Then, calculate the gpd solution required:

123 lb/day alum
5.30 lb alum p

Alum solution = = 23 gpd

er galsolution

The feeder setting, then, is 23 gpd alum solution. If the solution strength is expressed as a percent,
we use the following equation:

Chemical (mg/L) × Flow treated (MGD) × 8.34 lb/ gal

(2 4.101)

= Solution(mg/L)Solutionflow(MGD × ) × 8.34 lb/gal

■ EXAMPLE 24.73
Problem: The flow to a plant is 3.40 MGD. Jar testing indicates that the optimum alum dose is 10
mg/L. What should the gpd setting be for the solution feeder if the alum solution is a 52% solution?

Solution: A solution concentration of 52% is equivalent to 520,000 mg/L:

Desired dose(lb/day)=Actual dose(lb/day)

Chemical × Flow treated × 8.34 lb/gal = Solutio n × Solution flow × 8.34 lb/gal

10 mg/L × 3.40 MG D × 8.34 lb/gal = 520,00 mg/L × x MGD × 8.34 lb/ga l

x = ××
×

10 3.40 8.34

= 0.0000653 MGD

520,000 8.34