Chlorination
Dose mg/l - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You have been sent to a well site to inspect the Chlorinator. You find that the chlorine residual is 1.65 mg/l and the demand is 2.35 mg/l. What will the dose be?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (dose).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dose, mg/l = Demand, ppm or mg/l + Residual, ppm or mg/l (F)
THIRD. Follow the solution procedure below to solve the problem!
2.35 mg/l + 1.65 mg/l = 4.0 mg/l
Dose mg/l - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You are testing water samples in your water distribution system, you find that the chlorine (CL2) Residual is 0.35 ppm and your Dose is 2.0 ppm. What is the Demand?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (demand).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Demand, ppm or mg/l = Dose – Residual (F)
THIRD. Follow the solution procedure below to solve the problem!
2.0 ppm - 0.35 ppm = 1.65 ppm
Dose mg/l - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You are testing water samples in your water distribution system, you find that the chlorine (CL2) Demand is 2.5 mg/l and your Dose is 3.0 mg/l. What is the Residual?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (residual).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Residual, ppm or mg/l = Dose – Demand (F)
THIRD. Follow the solution procedure above to below the problem!
3.0 mg/l – 2.5 mg/l = 0.5 mg/l
Gas lbs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
How many pounds of chlorine gas will you need to disinfect a reservoir that contains 4 million gallons (MG) of water. The required dose is 12.5 mg/l.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (pounds, gas).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Gas, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)
THIRD. Follow the solution procedure below to solve the problem!
4 MG x 12.5 mg/l x 8.34 lbs/gal = 417 (Gas) lbs
Gas lbs - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A reservoir needs to be chlorinated. It is 35 ft tall, and the diameter is 55 ft. the dose is 15 mg/l. How many pounds of chlorine gas do you need to treat the reservoir?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (cylinder, gas).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)1
MG = Gallons divided by 1,000,000 (C&C)2
Gas, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)3
THIRD. Follow the solution procedure below to solve the problem!
.785 x 55 dia/ft x 55 dia/ft x 35 ft x 7.48 gal/ft3 = 621,676.83 gal
621,676.83 gal / 1,000,000 = .622 MG
.622 MG x 15mg/l x 8.34 lbs/gal = 77.81 lbs
Gas lbs - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Your water treatment plant has a basin that is 55 ft long 40 ft wide and 25 ft tall. The demand is 10 ppm and the residule is 5 ppm. How many pounds of chorine do you need to treat the water? Assume the basin is full.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (pounds, gas).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Cube, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (C&C)1
MG = Gallons divided by 1,000,000 (C&C)2
Dose, ppm or mg/l = Demand + Residual (C&C)3
Gas, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)4
THIRD. Follow the solution procedure below to solve the problem!
55 ft x 40 ft x 25 ft x 7.48 gal/ft3 = 411,400 gal
411,400 gal / 1,000,000 = 0.4114 MG
10 ppm + 5 ppm = 15 ppm
0.4114 MG x 15 ppm x 8.34 lbs/gal = 51.47 Lbs
Gas lbs - 4
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
How many pounds of Chlorine is required for a water main that is 12” dia and 2 miles long? The Demand is 8 ppm, the Residual is 2 ppm.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (pounds, gas).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Ft, Length = Mile, Length x 5,280 ft/mile (C&C)1
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)2
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)3
MG = Gallons divided by 1,000,000 (C&C)4
Dose, ppm or mg/l = Demand + Residual (C&C)5
Gas, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)6
THIRD. Follow the solution procedure below to solve the problem!
2 miles x 5,280 ft/mile = 10,560 ft
12 in / 12 in/ft = 1 dia/ft
0.785 x 1 dia/ft x 1 dia/ft x 10,560 ft x 7.48 gal/ft3 = 62,006.21 gal
62,006.21 gal / 1,000,000 = 0.062 MG
8 ppm + 2 ppm = 10 ppm
0.062 MG x 10 ppm x 8.34 lbs/gal = 5.17 lbs
Gas lbs - 5
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A reservoir is 25 ft tall and 40 ft in dia, the gauge at the bottom the tank says 6.50 PSI. you have to chlorinate it with a dose of 18 mg/l. How many pounds of chlorine do you need?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gas).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Head, ft = PSI x 2.31 ft/PSI (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)2
MG = Gallons divided by 1,000,000 (C&C)3
Gas, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)4
THIRD. Follow the solution procedure below to solve the problem!
6.50 PSI x 2.31 ft/PSI = 15 ft tall
.785 x 15 ft x 40 ft x 40 ft x 7.48 gal/ft3 = 140,923.20 gal
140,923.20 gal / 1,000,000 = .141 MG
.141 MG x 18 mg.l x 8.34 lbs/gal = 21.17 Lbs
HTH Solid lbs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You must disinfect a reservoir with 65% Calcium Hypochlorite (HTH). The tank has 3.5 MG of water and the dosage required is 15 mg/l. How many lbs. Of chlorine is needed?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (pounds, HTH).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Decimal = % divided by 100 (C&C)1
HTH Solids, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)2
% Strength / 100
THIRD. Follow the solution procedure below to solve the problem!
65% / 100 = .65
3.5 MG x 15 mg/l x 8.34 lbs/gal = 437.85 lbs/gal = 673.62 lbs
.65 .65
HTH Solid lbs - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
How many pounds of chlorine do you need to disinfect a basin that is 40 ft long 30 ft wide and 15 ft tall. The dosage is 18 mg/l with 65% strength.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (pounds, HTH).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Cube, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (C&C)1
MG = Gallons divided by 1,000,000 (C&C)2
Decimal = % divided by 100 (C&C)3
HTH Solids, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)4
% Strength / 100
THIRD. Follow the solution procedure below to solve the problem!
40 ft x 30 ft x 15 ft x 7.48 gal/ft = 134,640 gal
134,640 gal / 1,000,000 = 0.135 MG
65% / 100 = 0.65
0.135 MG x 18 mg/l x 8.34 lbs/gal = 20.27 lbs/gal = 31.18 lbs
0.65 0.65
HTH Solid lbs - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A 10 inch dia water main 5,500 ft long needs to be dosed with 63% Calcium hypochlorite solution. The Residule is 6mg/l. The Demand is 12 mg/l. How many pounds of Chlorine do you need?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (pounds, HTH).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)2
MG = Gallons divided by 1,000,000 (C&C)3
Dose, ppm or mg/l = Demand + Residual (C&C)4
Decimal = % divided by 100 (C&C)5
HTH Solids, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)6
% Strength / 100
THIRD. Follow the solution procedure below to solve the problem!
10 dia,in / 12 = 0.833 dia/ft
0.785 x 0.833 dia/ft x 0.833 dia/ft x 5,500 ft = 2,995.87 gal
2,995.87 gal / 1,000,000 = 0.00299 MG
12 mg/l + 6 mg/l = 18 mg/l
63% / 100 = 0.63
0.00299 MG x 18 mg/l x 8.34 lbs/gal = 0.449 lbs = 0.70 lbs
0.65 0.65
Liquid Gal - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A reservoir in your water system is holding 2 MG of water. How many gallons of 6.8% Sodium hypochlorite will be needed to disinfect the reservoir, the dosage required is 20 mg/l.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Decimal = % divided by 100 (C&C)1
Liquid, Gal = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)2
% Strength / 100 x Chemical Wt. lbs/gal
THIRD. Follow the solution procedure below to solve the problem!
6.8% / 100 = .068
2 MG x 20 mg/l x 8.34 lbs/gal = 333.6 lbs/MG = 588.4 Gal
0.068 x 8.34 lbs/gal 0.567 lbs
Liquid Gal - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
The water treatment plant tank measures 65 ft long, 45 ft wide, and 30 ft deep. The chlorine residule is 2.5 mg/l and the demand is 16.5 mg/l with a 25% strength, the chemical weight is 11.5 lbs/gal. How many gallons of Chlorine is needed to treat the tank?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Cube, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (C&C)1
MG = Gallons divided by 1,000,000 (C&C)2
Dose, ppm or mg/l = Demand + Residual (C&C)3
Decimal = % divided by 100 (C&C)4
Liquid, Gal = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)5
% Strength / 100 x Chemical Wt. lbs/gal
THIRD. Follow the solution procedure below to solve the problem!
65 ft x 45 ft x 30 ft x 7.48 gal/ft3 = 656,370 gal
656,370 gal / 1,000,000 = 0.656 MG
16.5 mg/l + 2.5 mg/l = 19 mg/l
25% / 100 = 0.25
0.656 MG x 19 mg/l x 8.34 lbs/gal = 103.95 lbs/gal = 36.09 gal
0.25 x 11.5 lbs/gal 2.88 lbs
Liquid Gal - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A water basin measures measures 25 ft long, 35 ft wide, and 20 ft tall. The dose is 17 ppm with a strength of 22%. The specific gravity is 1.5 how many gallons do you need to disinfect the water basin?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Cube, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (C&C)1
MG = Gallons divided by 1,000,000 (C&C)2
Decimal = % divided by 100 (C&C)3
Chemical Wt. (lbs/gal) = Specific Gravity x 8.34 lbs/gal (C&C)4
Liquid, Gal = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)5
% Strength / 100 x Chemical Wt. lbs/gal
THIRD. Follow the solution procedure below to solve the problem!
25 ft x 35 ft x 20 ft x 7.48 gal/ft3 = 130,900 gal
130,900 gal / 1,000,000 = 0.131 MG
22% / 100 = 0.22
1.5 x 8.34 lbs/gal = 12.51 lbs/gal
0.131 MG x 17 ppm x 8.34 lbs/gal = 18.57 lbs/gal = 7.30 gal
0.22 x 12.51 lbs/gal 2.57 lbs
Liquid Gal - 4
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You just installed 10,000 ft of 8 inch dia water main, and you have to treat it. You use chlorine at 15 ppm with 12.5% strength. How many gallons do you need?
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)2
MG = Gallons divided by 1,000,000 (C&C)3
Decimal = % divided by 100 (C&C)4
Liquid, Gal = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)5
% Strength / 100 x Chemical Wt. lbs/gal
THIRD. Follow the solution procedure below to solve the problem!
8 dia,in / 12 = 0.667 dia/ft
0.785 x 0.667 dia/ft x 0.667 dia/ft x 10,000 ft x 7.48 gal/ft3 = 26,122.9 gal
gal 26,122.9 / 1,000,000 = 0.0261 MG
12.5% / 100 = 0.125
0.0261 MG x 15 ppm x 8.34 lbs/gal = 3.27 lbs = 3.14 gal
0.125 x 8.34 lbs/gal 1.043 lbs
Liquid Gal - 5
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
The sample station on an 18 inch dia water main that is 2,000 ft long shows the residual is .5 mg/l and the demand is 15.5 mg/l, you wish to use 13.5% sodium hypochlorite, how many gallons will you need to dose the main with? The chemical weight is 12 lbs.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)2
MG = Gallons divided by 1,000,000 (C&C)3
Dose, ppm or mg/l = Demand + Residual (C&C)4
Decimal = % divided by 100 (C&C)5
Liquid, Gal = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)6
% Strength / 100 x Chemical Wt. lbs/gal
THIRD. Follow the solution procedure below to solve the problem!
18 dia/in / 12in/ft = 1.5 dia/ft
0.785 x 1.5 dia/ft x 1.5 dia/ft x 2,000 ft x 7.48 gal/ft3 = 26,423.1 gal
26,423.1 gal / 1,000,000 = 0.026 MG
15.5 mg/l + .5 mg/l = 16 mg/l
13.5 % / 100 = 0.135
0.026 MG x 16 mg/l x 8.34 lbs/gal = 3.47 lbs/gal = 2.14 lbs
0.135 x 12 lbs/gal 1.62 lbs/gal
Liquid Gal - 6
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
The sample station on an 18 inch dia water main that is 2,000 ft long shows the residual is .5 mg/l and the demand is 15.5 mg/l, you wish to use 13.5% sodium hypochlorite, how many gallons will you need to dose the main with? The chemical weight is 12 lbs.
SECOND. By using the KEY WORDS you see this is a (Chlorination) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)2
MG = Gallons divided by 1,000,000 (C&C)3
Dose, ppm or mg/l = Demand + Residual (C&C)4
Decimal = % divided by 100 (C&C)5
Liquid, Gal = Vol, MG x ppm or mg/l x 8.34 lbs/gal (F)6
% Strength / 100 x Chemical Wt. lbs/gal
THIRD. Follow the solution procedure below to solve the problem!
18 dia,in / 12in/ft = 1.5 dia/ft
0.785 x 1.5 dia/ft x 1.5 dia/ft x 2,000 ft x 7.48 gal/ft3 = 26,423.1 gal
26,423.1 gal / 1,000,000 = 0.026 MG
15.5 mg/l + .5 mg/l = 16 mg/l
13.5 % / 100 = 0.135
0.026 MG x 16 mg/l x 8.34 lbs/gal = 3.47 lbs/gal = 2.14 lbs
0.135 x 12 lbs/gal 1.62 lbs/gal