Geoscience Reference
In-Depth Information
23.4.4.5 Determining Chemical Solution Feeder Setting (Milliliters/Minute)
Some solution chemical feeders dispense chemical as milliliters per minute (mL/min). To calculate
the mL/min solution required, use the following procedure:
Solution (mL/min) = (gpd × 3785 mL/gal)/(1440 min/day)
(23.28)
EXAMPLE 23.31
Problem: The desired solution feed rate was calculated to be 9 gpd. What is this feed rate expressed
as m L/m in?
Solution:
Solution = (gpd × 3785 mL/gal)/(1440 min/day)
= (9 gpd × 3785 mL/gal)/(1440 min/day) = 24 mL/min
EXAMPLE 23.32
Problem: The desired solution feed rate has been calculated to be 25 gpd. What is this feed rate
expressed as mL/min?
Solution:
Solution = (gpd × 3785 mL/gal)/(1440 min/day)
= (25 gpd × 3785 mL/gal)/(1440 min/day) = 65.7 mL/min
Sometimes we will need to know the mL/min solution feed rate but we do not know the gpd
solution feed rate. In such cases, calculate the gpd solution feed rate first, using the following the
equation:
Chemical (mg/L)
× ×
hemical (lb) Solution (gal)
Flow (MGD)
8.34 lb/gal
gpd
=
(23.29)
C
23.4.5 d etermining p erCent s trength oF s olutions
The strength of a solution is a measure of the amount of chemical solute dissolved in the solution.
We use the following equation to determine the percent strength of a solution:
Chemical (lb)
Water (lb)
%Strength
=
×100
(23.30)
+
Chemical
(lb)
EXAMPLE 23.33
Problem: If a total of 10 oz. of dry polymer is added to 15 gal of water, what is the percent strength
(by weight) of the polymer solution?
Solution: Before calculating percent strength, the ounces of chemical must be converted to pounds
of chemical:
(10 oz.)/(16 oz./lb) = 0.625 lb chemical
 
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