water meter accuracy, percent solution strength and determining chlorine dosage solution strength in...

Water Meter Accuracy, Percent Solution Strength and Determining

Chlorine Dosage Solution Strength in Waterworks Operation

Water Meter Accuracy, Percent Solution Strength and Determining

Chlorine Dosage Solution Strength in Waterworks Operation

Math for Water TechnologyMTH 082

Lecture Chapter 6- Applied Math for Water Plant Operators

Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators

Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)

Math for Water TechnologyMTH 082

Lecture Chapter 6- Applied Math for Water Plant Operators

Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators

Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)

ObjectivesObjectives

1. Water meter accuracy

2. Percent solution strength

3. Learn to calculate lbs formula for different specific gravities of basic chemical solutions (percent strength)

1. Water meter accuracy

2. Percent solution strength

3. Learn to calculate lbs formula for different specific gravities of basic chemical solutions (percent strength)

Reading assignment: Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)

Reading assignment: Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)

Water Meter Accuracy Water Meter Accuracy Meter accuracy, %= (meter reading)* (100%)

Actual volume

IN % PROBLEMS UNITS MUST CANCEL!

(Gal and Gal, L and L)

Meter accuracy, %= (meter reading)* (100%)

Actual volume

IN % PROBLEMS UNITS MUST CANCEL!

(Gal and Gal, L and L)

•Given•Formula•Solve:


If a water meter is being tested in the lab and reads 2,412 L. A volumetric tank

shows that the actual value in the tank is 2,445 L. What is the percent accuracy of

the meter? ?

If a water meter is being tested in the lab and reads 2,412 L. A volumetric tank

shows that the actual value in the tank is 2,445 L. What is the percent accuracy of

the meter? ?

101.

40%

98.6

5%

0.01

%1.

35%

0% 0%0%

100%

V1= 2,412 L, V2=2,445 L

Meter accuracy, %= (meter reading)* (100%) Actual volume

MA%= (2,412L)(100%)/2,445 LMA%= 98.65%

V1= 2,412 L, V2=2,445 L


MA%= (2,412L)(100%)/2,445 LMA%= 98.65%

1. 101.4%

2. 98.65%

3. .010%

4. 1.35%

1. 101.4%

2. 98.65%

3. .010%

4. 1.35%



A water meter is being tested in the lab and it reads 500 ft3. A volumetric tank

shows that the actual value is 14,883 L. What is the percent accuracy of the

meter? ?

A water meter is being tested in the lab and it reads 500 ft3. A volumetric tank

shows that the actual value is 14,883 L. What is the percent accuracy of the

meter? ?

98.4

2%

101.

60%

96.6

6%

1.98

%

8%0%

77%

15%

V= 500 ft3 * 7.48 gal/1ft3 = 3740 gal,, V2=14,883 L and 1 Liter = 0.26 Gallons


Meter accuracy, %= (3740, gal)* (100%) 3870 gal

MA%= 96.66%

V= 500 ft3 * 7.48 gal/1ft3 = 3740 gal,, V2=14,883 L and 1 Liter = 0.26 Gallons


Meter accuracy, %= (3740, gal)* (100%) 3870 gal

MA%= 96.66%

1. 98.42%

2. 101.6%

3. 96.66%

4. 1.98%

1. 98.42%

2. 101.6%

3. 96.66%

4. 1.98%



If a meter had an accuracy of 99.1% and the reading on the meter indicated 1,500 L, what is the actual volume in liters?

If a meter had an accuracy of 99.1% and the reading on the meter indicated 1,500 L, what is the actual volume in liters?

148

6 L

15.

14 L

151

.4 L

151

3.6

L

21%

71%

0%7%

99.1% =.991 and 1,500 L 100% = 1.0


Actual volume, L= (Meter reading)* (100%) (Meter Accuracy, %)

AV= (1,500)(1.0)/(0.991) = 1513.6 LAV= 1513.6 L

99.1% =.991 and 1,500 L 100% = 1.0


Actual volume, L= (Meter reading)* (100%) (Meter Accuracy, %)

AV= (1,500)(1.0)/(0.991) = 1513.6 LAV= 1513.6 L

1. 1486 L

2. 15.14 L

3. 151.4 L

4. 1513.6 L

1. 1486 L

2. 15.14 L

3. 151.4 L

4. 1513.6 L

Percent Strength of a SolutionPercent Strength of a Solution% = Part *100 Whole

Amount of solute dissolved in solution!

% Strength = Chem lbs * 100 Total Sol, Lbs

% Strength = Chem lbs * 100

Water lb + Chem, Lbs


(Solution gal)(Water lb) + Chem, LbsNEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!

1 gram= 0.0022 lbs 1lb= 454 grams1 ounce = 0.0625 pounds

1 gal = 8.34 lbs (if water)

% = Part *100 Whole

Amount of solute dissolved in solution!

% Strength = Chem lbs * 100 Total Sol, Lbs


Water lb + Chem, Lbs


(Solution gal)(Water lb) + Chem, LbsNEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!



Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213)


•Given•Formula

•Solve:

•Given•Formula

•Solve:

If a total of 8 ounces of dry polymer are added to 10 gallons of water, what is the

percent strength (by weight) of the polymer solution?

If a total of 8 ounces of dry polymer are added to 10 gallons of water, what is the

percent strength (by weight) of the polymer solution?

6% 20%

0.05

%0.

60%

8%

77%

15%

0%

8 ounces of polymer= 1lb = 0.5 lbs 0.0625 oz% = Part *100 Whole

% Strength = Chem lbs *100 Water lb + Chem, Lbs

% Strength = 0.5 lbs *100

(10 gal) (8.34 lb/gal) + 0.5 lbs

% Strength = 0.5 lbs * 100% 83.9 lbs Sol

8 ounces of polymer= 1lb = 0.5 lbs 0.0625 oz% = Part *100 Whole



(10 gal) (8.34 lb/gal) + 0.5 lbs

% Strength = 0.5 lbs * 100% 83.9 lbs Sol 1. 6%

2. 20%

3. 0.05%

4. 0.6 %

1. 6%

2. 20%

3. 0.05%

4. 0.6 %

•Given•Formula

•Solve:

•Given•Formula

•Solve:

If 100 grams of dry polymer are dissolved in 5 gallons of water, what is the percent

strength (by weight) of the polymer solution?

If 100 grams of dry polymer are dissolved in 5 gallons of water, what is the percent

strength (by weight) of the polymer solution?

0.52

%5.

20%

2.41

%

24.6

3%

100%

0%0%0%

100 grams= 0.0022 lb = 0.22 lbs of chemical grams% = Part *100 Whole



(5 gal) (8.34 lb/gal) + 0.22 lbs


% strength =0.52%

100 grams= 0.0022 lb = 0.22 lbs of chemical grams% = Part *100 Whole



(5 gal) (8.34 lb/gal) + 0.22 lbs


% strength =0.52%

1. 0.52%

2. 5.2%

3. 2.41%

4. 24.63%

1. 0.52%

2. 5.2%

3. 2.41%

4. 24.63%

•Given•Formula

•Solve:

•Given•Formula

•Solve:

How many lbs of dry weight polymer due you need to add to a 25 gallon tank to make a 1%

solution?

How many lbs of dry weight polymer due you need to add to a 25 gallon tank to make a 1%

solution?

0.4

7 lb

s

2.1

lbs

208

lbs

47

lbs

18%

0%

9%

73%

25 gallons 1% solution

% = Part *100 Whole


1% = X lbs *100

(25 gal) (8.34 lb/gal) + X lbs

1 = 100X 208.5 lbs + X

208.5 lbs + 1X = 100 X208.5 lbs = 99 x208.5 lbs =X lbs 99

2.106 lbs = X

25 gallons 1% solution

% = Part *100 Whole


1% = X lbs *100

(25 gal) (8.34 lb/gal) + X lbs

1 = 100X 208.5 lbs + X

208.5 lbs + 1X = 100 X208.5 lbs = 99 x208.5 lbs =X lbs 99

2.106 lbs = X

1. 0.47 lbs

2. 2.1 lbs

3. 208 lbs

4. 47 lbs

1. 0.47 lbs

2. 2.1 lbs

3. 208 lbs

4. 47 lbs

Mixing Different Percent Strength Solutions

Mixing Different Percent Strength Solutions

Amount of solute dissolved in solution!% Strength of mixture = Chem lbs in Mix *100 Total lbs Sol in Mix

% Str of mix = Chem lbs in Sol #1 + Chem lbs in Sol #2 *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix

% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix

NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!1 gram= 0.0022 lbs 1lb= 454 grams1 ounce = 0.0625 pounds


Amount of solute dissolved in solution!% Strength of mixture = Chem lbs in Mix *100 Total lbs Sol in Mix

% Str of mix = Chem lbs in Sol #1 + Chem lbs in Sol #2 *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix


NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!1 gram= 0.0022 lbs 1lb= 454 grams1 ounce = 0.0625 pounds




•Given•Formula

•Solve:

•Given•Formula

•Solve:

If 20 lbs of a 10% strength solution are mixed with 50 lbs of a 1%

solution, what is the percent strength of the new solution?

If 20 lbs of a 10% strength solution are mixed with 50 lbs of a 1%

solution, what is the percent strength of the new solution?

37%

3.70

%0.

04%

0.07

%

0% 0%

15%

85%

Solution #1= 20 lbs of 10% Solution #2 = 50 lbs of 1%


% Str of mix = 20 lbs (10/100) + 50 lbs (1/100) *100 20 lbs Sol #1 + 50 lbs Sol #2 in Mix

% Str of mix = 2 lbs + 0.50 lbs *100

70 lbs Sol

% Str of mix = 3.7%

Solution #1= 20 lbs of 10% Solution #2 = 50 lbs of 1%


% Str of mix = 20 lbs (10/100) + 50 lbs (1/100) *100 20 lbs Sol #1 + 50 lbs Sol #2 in Mix

% Str of mix = 2 lbs + 0.50 lbs *100

70 lbs Sol

% Str of mix = 3.7%

1. 37%

2. 3.7 %

3. 0.036 %

4. .07%

1. 37%

2. 3.7 %

3. 0.036 %

4. .07%

•Given•Formula

•Solve:

•Given•Formula

•Solve:

If 5 gal of an 8% strength solution are mixed with 40 gal of a 0.5% solution, what is the

percent strength of the new solution? (Assume the 8% solution weighs 9.5 lbs/gal and the 0.5% solution weighs 8.34 lbs/gal)

If 5 gal of an 8% strength solution are mixed with 40 gal of a 0.5% solution, what is the

percent strength of the new solution? (Assume the 8% solution weighs 9.5 lbs/gal and the 0.5% solution weighs 8.34 lbs/gal)

0.04

%14

%1.

40%

1.70

%

9%0%

91%

0%

Solution #1= 5 gal of 8% Solution #2 = 40 gal of 0.5%


% Str of mix = (5 gal)(9.5lb/gal)(8/100) + 40 gal(8.34lb/gal)(0.5/100) *100 (5 gal)(9.5lb/gal) + 40 gal(8.34lb/gal)

% Str of mix = 3.8 lbs + 1.7 lbs *100

47.5 lbs Sol + 333.6 lbs

% Str of mix = 5.5 lbs *100 381.8 lbs

% Str of mix =1.4%

Solution #1= 5 gal of 8% Solution #2 = 40 gal of 0.5%


% Str of mix = (5 gal)(9.5lb/gal)(8/100) + 40 gal(8.34lb/gal)(0.5/100) *100 (5 gal)(9.5lb/gal) + 40 gal(8.34lb/gal)

% Str of mix = 3.8 lbs + 1.7 lbs *100

47.5 lbs Sol + 333.6 lbs

% Str of mix = 5.5 lbs *100 381.8 lbs

% Str of mix =1.4%

1. .04%

2. 14%

3. 1.4%

4. 1.7%

1. .04%

2. 14%

3. 1.4%

4. 1.7%

Different Percent Strength Solutions With Different

Gravities

Different Percent Strength Solutions With Different

Gravities

Lbs of polymer in liquid polymer = lbs of polymer in polymer solutions

(Liquid Polymer Lbs)(% str of liquid polymer/100) = (Polymer sol, lbs)(% str of polymer Sol/100)

Liquid Polymer Lbs)((density)(% str of liquid polymer/100) = (Polymer sol, lbs)(density)(% str of polymer Sol/100

Lbs of polymer in liquid polymer = lbs of polymer in polymer solutions

(Liquid Polymer Lbs)(% str of liquid polymer/100) = (Polymer sol, lbs)(% str of polymer Sol/100)

Liquid Polymer Lbs)((density)(% str of liquid polymer/100) = (Polymer sol, lbs)(density)(% str of polymer Sol/100

Chapter 9- Applied Math for Water Plant Operators pg. 210-211.Chapter 9- Applied Math for Water Plant Operators pg. 210-211.

•Given•Formula

•Solve:

•Given•Formula

•Solve:

A 10% liquid polymer is to be used in making up a polymer solution. How many lbs of liquid

polymer should be mixed with water to

produce 130 lbs of a 0.6% polymer solution?

A 10% liquid polymer is to be used in making up a polymer solution. How many lbs of liquid

polymer should be mixed with water to

produce 130 lbs of a 0.6% polymer solution?

7.8

lbs

130

0 lb

s

0.7

8 lb

s

13

lbs

83%

8%8%0%

10% Liquid polymer ?Lbs of liquid polymer =130 lbs of 0.6% polymer solution----Lbs of polymer in liquid polymer = lbs of polymer in polymer solutions

(Liquid Polymer Lbs)(% str of liquid polymer/100) = (Polymer sol, lbs)(% str of polymer sol/100)

(X lbs)(10/100) = (130 lbs)(0.6l/100)

(X lbs)(0.1) = (0.78 lbs)(X lbs) = (0.78 lbs)/0.1X lbs = 7.8 lbs

10% Liquid polymer ?Lbs of liquid polymer =130 lbs of 0.6% polymer solution----Lbs of polymer in liquid polymer = lbs of polymer in polymer solutions

(Liquid Polymer Lbs)(% str of liquid polymer/100) = (Polymer sol, lbs)(% str of polymer sol/100)

(X lbs)(10/100) = (130 lbs)(0.6l/100)

(X lbs)(0.1) = (0.78 lbs)(X lbs) = (0.78 lbs)/0.1X lbs = 7.8 lbs

1. 7.8 lbs

2. 1300 lbs

3. 0.78 lbs

4. 13 lbs

1. 7.8 lbs

2. 1300 lbs

3. 0.78 lbs

4. 13 lbs

•Given•Formula

•Solve:

•Given•Formula

•Solve:

How many gallons of a 9% liquid polymer should be mixed with water to produce 50 gallons of a 0.5% polymer solution? The

density of the polymer liquid is 10.34 lb/gal, assume the density of the polymer solution is

8.34 lb/gal?

How many gallons of a 9% liquid polymer should be mixed with water to produce 50 gallons of a 0.5% polymer solution? The

density of the polymer liquid is 10.34 lb/gal, assume the density of the polymer solution is

8.34 lb/gal?

1 g

al

1.9

5 gal

2.2

gal

22

gal

9%18%

73%

0%

9% Liquid polymer ?gallons of liquid polymer =50 gal of 0.5% polymer solution---- Density of polymer 10.4 lb/gal Density of polymer solution= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions

(Liquid Polymer gal)(D)(% str of liquid polymer/100) = (Polymer sol, gal)(D)(% str pol sol/100)(Liquid Polymer gal?)(10.34 lb/gal)(9/100) = (50 gal)(8.34 lb/gal)(0.5/100)(Liquid Polymer gal?)(10.34 lb/gal)(9/100) = (50 gal)(8.34 lb/gal)(0.5/100)(Liquid Polymer gal?)(0.936 lb/gal) = 2.085 lbs(Liquid Polymer gal?)(= 2.085 lbs/(0.936 lb/gal) (Liquid Polymer gal?)= 2.22 gal

9% Liquid polymer ?gallons of liquid polymer =50 gal of 0.5% polymer solution---- Density of polymer 10.4 lb/gal Density of polymer solution= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions

(Liquid Polymer gal)(D)(% str of liquid polymer/100) = (Polymer sol, gal)(D)(% str pol sol/100)(Liquid Polymer gal?)(10.34 lb/gal)(9/100) = (50 gal)(8.34 lb/gal)(0.5/100)(Liquid Polymer gal?)(10.34 lb/gal)(9/100) = (50 gal)(8.34 lb/gal)(0.5/100)(Liquid Polymer gal?)(0.936 lb/gal) = 2.085 lbs(Liquid Polymer gal?)(= 2.085 lbs/(0.936 lb/gal) (Liquid Polymer gal?)= 2.22 gal

1. 1 gal

2. 1.95 gal

3. 2.2 gal

4. 22 gal

1. 1 gal

2. 1.95 gal

3. 2.2 gal

4. 22 gal

•Given•Formula

•Solve:

•Given•Formula

•Solve:

A 10% liquid polymer will be used in making up a solution. How many gallons of liquid polymer should be added to the water to make up 40 gallons of 0.35% polymer solution. The liquid polymer has a specific gravity of 1.1. Assume the polymer has a specific

gravity of 1.0?

A 10% liquid polymer will be used in making up a solution. How many gallons of liquid polymer should be added to the water to make up 40 gallons of 0.35% polymer solution. The liquid polymer has a specific gravity of 1.1. Assume the polymer has a specific

gravity of 1.0?

1.1

gal

0.9

17 g

al

1.3

gal

1.0

6 gal

0%

10%

80%

10%

10% Liquid polymer ?gallons of liquid polymer =40 gal of 0.35% polymer solution---- liquid polymer specific gravity= 1.1, polymer solution specific gravity =11.1 (8.34 lb/gal) = 9.17 lb/gal polymer solution 1(8.34 lb/gal)= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions(Liquid Polymer gal)(D)(%/100) = (Polymer sol, lbs)(D)(% str pol sol/100)

(Liquid Polymer gal)(9.17 lb/gal)(10/100) = (40 gal)(8.34 lb/gal)(0.35/100)(Liquid Polymer gal?)(0.917 lb/gal) = (333.6 lb)(0.35/100)(Liquid Polymer gal?)(0.917 lb/gal) = (333.6 lb)(0.35/100)(Liquid Polymer gal?)(0.917 lb/gal) = 1.16 lbs(Liquid Polymer gal?)(= 1.16 lbs/(0.917 lb/gal) (Liquid Polymer gal?)= 1.27 gal

10% Liquid polymer ?gallons of liquid polymer =40 gal of 0.35% polymer solution---- liquid polymer specific gravity= 1.1, polymer solution specific gravity =11.1 (8.34 lb/gal) = 9.17 lb/gal polymer solution 1(8.34 lb/gal)= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions(Liquid Polymer gal)(D)(%/100) = (Polymer sol, lbs)(D)(% str pol sol/100)

(Liquid Polymer gal)(9.17 lb/gal)(10/100) = (40 gal)(8.34 lb/gal)(0.35/100)(Liquid Polymer gal?)(0.917 lb/gal) = (333.6 lb)(0.35/100)(Liquid Polymer gal?)(0.917 lb/gal) = (333.6 lb)(0.35/100)(Liquid Polymer gal?)(0.917 lb/gal) = 1.16 lbs(Liquid Polymer gal?)(= 1.16 lbs/(0.917 lb/gal) (Liquid Polymer gal?)= 1.27 gal1. 1.1 gal

2. 0.917 gal

3. 1.3 gal

4. 1.06 gal

1. 1.1 gal

2. 0.917 gal

3. 1.3 gal

4. 1.06 gal

•Given•Formula

•Solve:

•Given•Formula

•Solve:

If 12 gallons of a 10% strength solution are added to 48 gallons of 0.6% strength solution, what is the

percent strength of the solution mixture? (Assume the 10% strength solution weighs 10.12 lbs/gal and

the 0.6% solution weighs 8.7 lbs /gal)?

If 12 gallons of a 10% strength solution are added to 48 gallons of 0.6% strength solution, what is the

percent strength of the solution mixture? (Assume the 10% strength solution weighs 10.12 lbs/gal and

the 0.6% solution weighs 8.7 lbs /gal)?

1% 25%

36.7

0%

2.70

%

0%

64%

9%

27%

Sol#1=12 gallons of a 10% solution Sol #2= 48 gal of 0.6%solution---- D=10.12 lbs/gal D2=8.7 lbs/gal

% Str of mix = (Chem lbs in Sol #1)(D) (% str/100) + (Chem lbs in Sol#2)(D) (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix % Str of mix = (12 gal)(10.12 lb/gal)(10/100) + (48 gal)(8.7 lb/gal)(0.6/100) * 100 (12 gal)(10.12 lb/gal) + (48 gal)(8.7 lb/gal)

% Str of mix = (12.2 lb) + (2.5 lbs) * 100 (122.4 lbs) + (417.6 lbs)

% Str of mix = 14.7 lbs * 100 540 lbs

% Str of mix =2.7 %

Sol#1=12 gallons of a 10% solution Sol #2= 48 gal of 0.6%solution---- D=10.12 lbs/gal D2=8.7 lbs/gal

% Str of mix = (Chem lbs in Sol #1)(D) (% str/100) + (Chem lbs in Sol#2)(D) (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix % Str of mix = (12 gal)(10.12 lb/gal)(10/100) + (48 gal)(8.7 lb/gal)(0.6/100) * 100 (12 gal)(10.12 lb/gal) + (48 gal)(8.7 lb/gal)

% Str of mix = (12.2 lb) + (2.5 lbs) * 100 (122.4 lbs) + (417.6 lbs)

% Str of mix = 14.7 lbs * 100 540 lbs

% Str of mix =2.7 %1. 1%

2. 25%

3. 36.7%

4. 2.7 %

1. 1%

2. 25%

3. 36.7%

4. 2.7 %

Different Chlorine Percent Strength Solutions

Different Chlorine Percent Strength Solutions

% Cl2 strength= (Hyp Lbs)(% available hyp/100) water lbs + hypo lbs% Cl2 strength= (Hyp Lbs)(% available hyp/100) water lbs + hypo lbs

Chapter 12- Applied Math for Water Plant Operators pg. 307-313.Chapter 12- Applied Math for Water Plant Operators pg. 307-313.

NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!



•Given•Formula

•Solve:

•Given•Formula

•Solve:

If a total of 64 ounces of calcium hypochlorite (65% available chlorine) are added to 10

gallons of water, what is the percent chlorine strength (by weight) of the solution?

If a total of 64 ounces of calcium hypochlorite (65% available chlorine) are added to 10

gallons of water, what is the percent chlorine strength (by weight) of the solution?

5% 3% 18% 6%

25% 25%25%25%

64 ounces * 0.0625 lbs/1 oz = 4 lbs

% chlorine = Hypo lbs (% Hypo str/100) *100 water lbs + Hyo lbs

% chlorine = (4 lbs)(65/100) *100 (10 gal)(8.34 lb/gal) + 4 lbs

% chlorine = 4lbs (.65) *100

83.4lbs + 4 lbs

% chlorine = 2.6 lbs *100 87.4 lbs

% chlorine =3.0 %

64 ounces * 0.0625 lbs/1 oz = 4 lbs

% chlorine = Hypo lbs (% Hypo str/100) *100 water lbs + Hyo lbs

% chlorine = (4 lbs)(65/100) *100 (10 gal)(8.34 lb/gal) + 4 lbs

% chlorine = 4lbs (.65) *100

83.4lbs + 4 lbs

% chlorine = 2.6 lbs *100 87.4 lbs

% chlorine =3.0 %

1. 5%

2. 3%

3. 18%

4. 6%

1. 5%

2. 3%

3. 18%

4. 6%

Today’s objective: Filter Loading Rates, Filter Backwash Rates has been met?

Today’s objective: Filter Loading Rates, Filter Backwash Rates has been met?

Stro

ngly A

gree

Agre

e

Neu

tral

Dis

agre

e

Stro

ngly D

isag

ree

20% 20% 20%20%20%

1. Strongly Agree

2. Agree

3. Neutral

4. Disagree

5. Strongly Disagree

1. Strongly Agree

2. Agree

3. Neutral

4. Disagree

5. Strongly Disagree

Team ScoresTeam Scores

110.7 left side

66.95 right side

water meter accuracy, percent solution strength and determining chlorine dosage solution strength in...

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