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Math for Water Technology MTH 082 Lecture Chapter 6- Applied Math for Water Plant Operators

Water Meter Accuracy, Percent Solution Strength and Determining Chlorine Dosage Solution Strength in Waterworks Operation. Math for Water Technology MTH 082 Lecture Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117)

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Math for Water Technology MTH 082 Lecture Chapter 6- Applied Math for Water Plant Operators

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  1. Water Meter Accuracy, Percent Solution Strength and Determining Chlorine Dosage Solution Strength in Waterworks Operation Math for Water Technology MTH 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)

  2. Objectives 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 • Percent solution strength • Learn to calculate lbs formula for different specific gravities of basic chemical solutions (percent strength)

  3. Water Meter Accuracy Meter accuracy, %= (meter reading)* (100%) Actual volume IN % PROBLEMS UNITS MUST CANCEL! (Gal and Gal, L and L)

  4. 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? ? • 101.4% • 98.65% • .010% • 1.35% • Given • Formula • Solve: V1= 2,412 L, V2=2,445 L Meter accuracy, %= (meter reading)* (100%) Actual volume MA%= (2,412L)(100%)/2,445 L MA%= 98.65%

  5. 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? ? • 98.42% • 101.6% • 96.66% • 1.98% • Given • Formula • Solve: V= 500 ft3 * 7.48 gal/1ft3 = 3740 gal,, V2=14,883 L and 1 Liter= 0.26 Gallons Meter accuracy, %= (meter reading)* (100%) Actual volume Meter accuracy, %= (3740, gal)* (100%) 3870 gal MA%= 96.66%

  6. 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? • 1486 L • 15.14 L • 151.4 L • 1513.6 L 99.1% =.991 and 1,500 L 100% = 1.0 Meter accuracy, %= (meter reading)* (100%) Actual volume Actual volume, L= (Meter reading)* (100%) (Meter Accuracy, %) AV= (1,500)(1.0)/(0.991) = 1513.6 L AV= 1513.6 L • Given • Formula • Solve:

  7. Percent Strength of a Solution 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) % = Part *100 Whole Amount of solute dissolved in solution! % Strength = Chem lbs * 100 Total Sol, Lbs % Strength = Chem lbs * 100 Water lb + Chem, Lbs % Strength = Chem lbs * 100 (Solution gal)(Water lb) + Chem, Lbs NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!! 1 gram= 0.0022 lbs 1lb= 454 grams 1 ounce = 0.0625 pounds 1 gal = 8.34 lbs (if water)

  8. 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% • 0.6 % • Given • Formula • Solve: 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

  9. 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% • 5.2% • 2.41% • 24.63% • Given • Formula • Solve: 100 grams= 0.0022 lb = 0.22 lbs of chemical grams % = Part *100 Whole % Strength = Chem lbs *100 Water lb + Chem, Lbs % Strength = 0.22 lbs *100 (5 gal) (8.34 lb/gal) + 0.22 lbs % Strength = 0.22 lbs * 100% 41.92 lbs Sol % strength =0.52%

  10. How many lbs of dry weight polymer due you need to add to a 25 gallon tank to make a 1% solution? • 0.47 lbs • 2.1 lbs • 208 lbs • 47 lbs • Given • Formula • Solve: 25 gallons 1% solution % = Part *100 Whole % Strength = Chem lbs *100 Water lb + Chem, Lbs 1% = X lbs *100 (25 gal) (8.34 lb/gal) + X lbs 1 = 100X 208.5 lbs + X 208.5 lbs + 1X = 100 X 208.5 lbs = 99 x 208.5 lbs =X lbs 99 2.106 lbs = X

  11. Mixing Different Percent Strength Solutions 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) 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 grams 1 ounce = 0.0625 pounds 1 gal = 8.34 lbs (if water)

  12. 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% • 3.7 % • 0.036 % • .07% • Given • Formula • Solve: Solution #1= 20 lbs of 10% Solution #2 = 50 lbs of 1% % 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 % 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%

  13. 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) • .04% • 14% • 1.4% • 1.7% • Given • Formula • Solve: Solution #1= 5 gal of 8% Solution #2 = 40 gal of 0.5% % 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 % 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%

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

  15. 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 • 1300 lbs • 0.78 lbs • 13 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)/0.1 X lbs = 7.8 lbs • Given • Formula • Solve:

  16. 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 gal • 1.95 gal • 2.2 gal • 22 gal • Given • Formula • Solve: 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/gal Lbs 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

  17. 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 • 0.917 gal • 1.3 gal • 1.06 gal • Given • Formula • Solve: 10% Liquid polymer ?gallons of liquid polymer =40 gal of 0.35% polymer solution---- liquid polymer specific gravity= 1.1, polymer solution specific gravity =1 1.1 (8.34 lb/gal) = 9.17 lb/gal polymer solution 1(8.34 lb/gal)= 8.34 lb/gal Lbs 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

  18. 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% • 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 % • Given • Formula • Solve:

  19. Different Chlorine Percent Strength Solutions Chapter 12- Applied Math for Water Plant Operators pg. 307-313. % Cl2 strength= (Hyp Lbs)(% available hyp/100) water lbs + hypo lbs NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!! 1 gram= 0.0022 lbs 1lb= 454 grams 1 ounce = 0.0625 pounds 1 gal = 8.34 lbs (if water)

  20. 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% • Given • Formula • Solve: 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 %

  21. Today’s objective: Filter Loading Rates, Filter Backwash Rates has been met? • Strongly Agree • Agree • Neutral • Disagree • Strongly Disagree

  22. Team Scores

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