Water Treatment Processes - Mathematics and Statistics for the Environment - page 618

Now calculate percent strength:

Chemical (lb)
Water (lb)

%Strength = × 100

+ Chemical (lb)

0.625 lb chemical
(15gal

0.625 lb chemical
125.7

= ×= 100 ×= 100 0.%

× 8.34 lb /gal) + 0.625 lb lb solution

■ EXAMPLE 23.34
Problem: If 90 g (1 g = 0.0022 lb) of dry polymer is dissolved in 6 gal of water, what percent
strength is the solution?

Solution: First, convert grams of chemical to pounds of chemical:

90 g polymer × 0.0022 lb/g = 0.198 lb polymer

Now calculate percent strength of the solution:

lb Polymer
lb Water

%Strength = ×100

+ lb Polymer

0.198 lb Polymer

= ×=% 100 4

(6 gal × 8.34 lb/gal) + 01 . 98 lb Polymer

23.4.5.1 Determining Percent Strength of Liquid Solutions
When using liquid chemicals to make up solutions (e.g., liquid polymer), a different calculation is
required, as shown below:

Liquid polymer (lb) × Liquid polymer (% stren gth)

100
Polymersolution(lb)Polymer solu

(23.31)

× tion (% strength)

=

100

■ EXAMPLE 23.35
Problem: A 12% liquid polymer is to be used in making up a polymer solution. How many pounds
of liquid polymer should be mixed with water to produce 120 lb of a 0.5% polymer solution?

Solution:

Liq. polymer (lb) × Liq. polymer (% strength) Polymersol.(lb)Polymer sol. (% stre × ngth)

=

100 100

x lb × 12

120 lb × 05 .

=

100 100

120 .

0012

× 0 005

x =

.
x = 5lb