Basic Math Midterm Review Math for Water Technology MTH Math for Water Technology MTH 082-2

RULES FOR MIDTERM 1.DRAW AND LABEL DIAGRAM 2.LIST FORMULAS 3 SOLVE EACH FORMULA INDIVIDUALLY 4. ISOLATE THE PARAMETERS NECESSARY 5. USE YOUR UNITS TO GUIDE YOU 6. SOLVE THE PROBLEM 1.DRAW AND LABEL DIAGRAM 2.LIST FORMULAS 3 SOLVE EACH FORMULA INDIVIDUALLY 4. ISOLATE THE PARAMETERS NECESSARY 5. USE YOUR UNITS TO GUIDE YOU 6. SOLVE THE PROBLEM

Chlorine Concentrations Hypochlorite or Available Purity = DIVIDE 1.Sodium hypochlorite 5 to 15% available chlorine 2. Calcium hypochlorite (HTH) 65-70% available chlorine 3. Chlorine gas 100% available chlorine or 100% active chlorine 1.Sodium hypochlorite 5 to 15% available chlorine 2. Calcium hypochlorite (HTH) 65-70% available chlorine 3. Chlorine gas 100% available chlorine or 100% active chlorine

% Dry Strength of Solution 1. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Solution(lbs) 1. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Solution(lbs) 2. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Water, Lbs + Chlorine(lbs) 2. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Water, Lbs + Chlorine(lbs) 3. Dry chlorine % Chlorine Strength= Hypo(lbs)(% available Cl) X Water, Lbs+ hypo(lbs) (% available Cl) Dry chlorine % Chlorine Strength= Hypo(lbs)(% available Cl) X Water, Lbs+ hypo(lbs) (% available Cl) 100

Dry Hypochlorite Feed Rate 1.Hypochlorite (lb/day)=lbs/day chlorine % available chlorine Hypochlorite (lb/day)= (mg/L chlorine)(MGD)(8.34) %strength hypochl Hypochlorite (lbs)= (mg/L chlorine)(MG tank Vol)(8.34) %strength hypochl Hypochlorite (lb/day)=lbs/day chlorine % available chlorine Hypochlorite (lb/day)= (mg/L chlorine)(MGD)(8.34) %strength hypochl Hypochlorite (lbs)= (mg/L chlorine)(MG tank Vol)(8.34) %strength hypochl 100

% Liquid Strength of Solution 1. Liquid chlorine Lbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution 1. Liquid chlorine Lbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution 2. Liquid chlorine (liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp) Liquid chlorine (liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp) Liquid chlorine (liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp) Liquid chlorine (liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp)

Lbs/day When do you multiply? Available to Active (multiply) Twenty 20 pounds of 50% available chlorine equals 10 lbs pounds of 100% active chlorine ? 20 lbs 10 lbs Active is pure Available is unpure 20 lbs (0.50)= 10 lbs

Lbs/day When do you divide? Active to Available Forty 40 pounds of 100% active chlorine (chlorine gas) has the same disinfection power as 61 lbs pounds of 65% available HTH chlorine ? 40 lbs 61 lbs Active is pure Available is unpure 40 lbs = 61 lbs (.65)

Lbs/day When do you multiply? Available to Active (multiply) HW #4 Problem 3 Twenty two 22 pounds of 65% available chlorine equals 14.3 lbs pounds of 100% active chlorine! HW #4 Problem 3 Twenty two 22 pounds of 65% available chlorine equals 14.3 lbs pounds of 100% active chlorine! 22 lbs 14.3 lbs Active is pure Available is unpure 22 lbs (0.65)= 14.3 lbs

Lbs/day When do you multiply? Available to Available (multiply) HW #3 Problem 3 How many pounds of available chlorine are in 2 pounds of 65% powdered dry (available) chlorine? HW #3 Problem 3 How many pounds of available chlorine are in 2 pounds of 65% powdered dry (available) chlorine? 2 lbs of 65% ava 1.3 lbs is available Available is unpure 2 lbs (0.65)= 1.3 lbs

Lbs/day When do you divide? When do you multiply? PURITY IS THE KEY!!!!! If you have a certain amount of <100% chlorine (available) and you want to know how much pure (active chlorine) that is you multiply!!!! (Need less of a more pure substance) When using hypochlorite, more lbs/day must be used because it is not pure! Hypochlorite is “available” chlorine If you have a certain # of lbs of 100% pure chlorine (active) and what you need is a certain percentage (available) then you divide (its not pure so you will need more of it!)

Chlorine in lbs DOSE in a tank! 1.Wells are disinfected 2.Tanks and reservoirs get chlorinated 3.Pipelines get chlorinated after repair or installation 1.Wells are disinfected 2.Tanks and reservoirs get chlorinated 3.Pipelines get chlorinated after repair or installation Example 5: A storage tank needs to be disinfected with a 50 mg/L chlorine solution. If the tank holds 70,000 gallons, how many lbs of chlorine gas will be needed? Step 1 Chlorine mg/L to lbs/day formula

Hypochlorite DOSAGE 1.Calculate lbs/day chlorine 2. Calculate lbs/day hypochlorite needed by dividing the lbs/day chlorine by the percent available chlorine 1.Calculate lbs/day chlorine 2. Calculate lbs/day hypochlorite needed by dividing the lbs/day chlorine by the percent available chlorine Example 6: A wastewater flow of 85,000 gpd requires a chlorine dose of 25 mg/L. If sodium hypochlorite (15% available chlorine) is used, how many lbs/day of sodium hypochlorite are required? How many gal/day is this? Step 2 Calculate lbs/day sodium hypochlorite (1180 lbs/d)/(8.34 lbs/gal)= 141 gal/day sodium hypochlorite

Determining Cl Concentrations from Hypochlorite dosage 1.Disinfection requires 280 lb/day chlorine. If CaOCl (65% available Cl) is used how many lbs day are required 1.Disinfection requires 280 lb/day chlorine. If CaOCl (65% available Cl) is used how many lbs day are required (%concentration)(X lb/day)= total lbs/day required Rearrange: (X lb /day) = Total lbs/d = (280 lbs) = lb/d CaOCL % concentration (.65) (%concentration)(X lb/day)= total lbs/day required Rearrange: (X lb /day) = Total lbs/d = (280 lbs) = lb/d CaOCL % concentration (.65)

Solutions with Specific Gravity 1.How many pounds of available chlorine are in 5 gallons of 12% hypochlorite solution if its specific gravity is 1.19? Lb=(% concentration) X (Specific gravity) X (lb/gal) X (gal) Lb= (.12) X (1.19) X (8.34 lb/gal) X (5 gal) Lb=5.95 lb Lb=(% concentration) X (Specific gravity) X (lb/gal) X (gal) Lb= (.12) X (1.19) X (8.34 lb/gal) X (5 gal) Lb=5.95 lb Calcium hypochlorite, gal= (lb 100%)/(8.34 lb/gal)(CaOCl2 %) Calcium hypochlorite, lbs= (gal CaOCl2)(8.34 lb/gal)(62.5%)/(100%)

Hypochlorite Solutions is disinfected with 62.5% calcium hypochlorite solution, If 3.03 gal of CaOCl 2 were used, what must have been the dosage of the calcium hypochlorite? Assume the calcium hypochlorite weighs 8.34 lb/gal. Get answer in (lbs) to solve for dosage in lbs formula! Calcium hypochlorite, lbs= (gal CaOCl 2 )(8.34 lb/gal)( % Available) (100%) Calcium hypochlorite, lbs= (gal CaOCl 2 )(8.34 lb/gal)( % Available) (100%) Calcium hypochlorite, lbs= lbs Calcium hypochlorite, lbs= lbs Calcium hypochlorite, lbs= (3.03 gal CaOCl 2 )(8.34 lb/gal)(.625) Calcium hypochlorite, lbs= (3.03 gal CaOCl 2 )(8.34 lb/gal)(.625) Solve lbs formula Number of lb=(MG)(Dosage)(8.34 lb/gal) Solve lbs formula Number of lb=(MG)(Dosage)(8.34 lb/gal) Dosage mg/L CaOCl 2 =15.79 lbs CaOCl 2 /(0.038MG)(8.34 lb/gal)

Calcium Hypochlorite Solutions Midterm 3. storage tank requires disinfection with a 65% calcium hypochlorite solution. How many gallons of hypochlorite solution are required if the desired dose is 50.0 mg/L. Assume the calcium hypochlorite solution weighs lb/gal.? Calcium hypochlorite, gal= (lb CaOCl 100%) (10.21 lb/gal)(CaOCl 2 % available) Calcium hypochlorite, gal= (51.7 lb CaOCl2) = 7.79 gal (10.21 lb/gal)(.65) Calcium hypochlorite, gal= (lb CaOCl 100%) (10.21 lb/gal)(CaOCl 2 % available) Calcium hypochlorite, gal= (51.7 lb CaOCl2) = 7.79 gal (10.21 lb/gal)(.65)

Specific Gravity, LBS, Gallons, Solution Strength

Problem #1 1.Determine the chlorinator setting (lbs/day) needed to treat a flow of 3.2 MGD with a chlorine dose of 1.7 mg/L (5 pts).

Problem #2 2. The chorine demand was calculated to be 11 lb/day, and the residual was measured at 0.5 mg/L. What was the chlorine dose (lbs) that day, if the flow was 1.2 MGD?

Problem #3 3. How many gallons of 15% liquid chorine (s.g. = 1.2) would be required to dose a circular tank to a 50 mg/L concentration? The tank diameter is 25 ft. and the water depth in the tank is 5 ft. ? Gallons 1.2 SG 15% OCl 1.2 SG 15% OCl 25 ft 5 ft Volume= 0.785(diameter 2 )(depth) Volume= 0.785(25ft) 2 (5 ft) Volume=2453ft 3 Conversion= 2453 ft 3 (7.48 gal/1ft 3 ) Conversion=18349 gallons Conversion= MG Volume= 0.785(diameter 2 )(depth) Volume= 0.785(25ft) 2 (5 ft) Volume=2453ft 3 Conversion= 2453 ft 3 (7.48 gal/1ft 3 ) Conversion=18349 gallons Conversion= MG

Problem #4 4. A barrel contains 30 gallons of 12% liquid hypochlorite at a specific gravity of How many pounds of dry HTH (65% active hypochlorite) would be needed to replace the chlorine in the barrel? 1.19 SG 12% OCl= 35 lbs pure Cl SG 12% OCl= 35 lbs pure Cl2 (%concentration)(X lb)= total lbs required X lb= lbs/.65 X=54.96 lbs of 65% HTH The lbs/day formula calculates the pure concentration of chlorine within a % solution of chlorine. Thus in the 12% hypochlorite solution there are 35 lbs of pure Cl (A). Imagine the same size barrel of a higher concentration of OCl- as 65% HTH (B). Which solution has more lbs of pure Cl 2 ? Well, the 65% it’s a more concentrated solution so there will be more pure chlorine in it. (%concentration)(X lb)= total lbs required X lb= lbs/.65 X=54.96 lbs of 65% HTH The lbs/day formula calculates the pure concentration of chlorine within a % solution of chlorine. Thus in the 12% hypochlorite solution there are 35 lbs of pure Cl (A). Imagine the same size barrel of a higher concentration of OCl- as 65% HTH (B). Which solution has more lbs of pure Cl 2 ? Well, the 65% it’s a more concentrated solution so there will be more pure chlorine in it. 65% OCl as HTH 55 lbs pure Cl2 65% OCl as HTH 55 lbs pure Cl2 A B