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Jar Testing of Chemical Dosages Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit.

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Presentation on theme: "Jar Testing of Chemical Dosages Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit."— Presentation transcript:

1 Jar Testing of Chemical Dosages Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit

2 Jar Testing Determination of most effective chemical Determination of most effective dosage Determination of optimum point of application Evaluation of polymers

3 Jar Testing Equipment Needed: Gang Stirrer 6 Graduated Beakers, 1500 ml 2 Graduated Pipets, 10 ml 1 Graduated Cylinder, 1000 ml Scale for weighing chemicals Analytical Equipment

4 Jar Testing Preparation of Stock Solutions For alum, lime, other dry materials: Use a 1 % solution. Dissolve 10 grams into 1000 ml distilled water. (1 ml = 10 mg/l in 1000 ml)

5 For Ferric Chloride, other liquid materials: Use a 1 % solution. Obtain % solution and specific gravity from supplier. Dilute appropriate volume up to 1000 ml to make a 1 % (10,000 mg/L) solution. (1ml=10 mg/l in 1000 ml) Preparation of Stock Solutions

6 Solution Dilution 45 % X 1.49 X V 1 = 1 % X 1000 ml C 1 = 45 % (weighs 1.49 grams/ml) V 1 = ? ml C 2 = 1 % (weighs 1.00 grams/ml V 2 = 1000 ml Have 45 % Ferric Chloride Solution (Specific Gravity 1.49) Need 1000 mls of a 1 % (10,000 mg/L) Sol. C 1 X V 1 = C 2 X V 2 = 14.91 ml 1 % X 1000 ml 45 % X 1.49 V1 =V1 =

7 Jar Testing For Dry or Liquid Polymers: Use a 0.01 % (100 mg/L) solution. Weigh 0.1 gram and dissolve in 1000 ml distilled water. (1 ml = 0.1 mg/l in 1000 ml) Preparation of Stock Solutions

8 Blank 5 mg/L 10 mg/L 15 mg/L 20 mg/L 25 mg/L Calculate the volume of a 1 % Ferric Chloride solution that would be added to get the dosage required. 1000 ml 1% = 10,000 mg/L 10,000 mg/L X ? ml = 5 mg/L X 1000 ml ? ml = 5 mg/L X 1000 ml 10,000 mg/L ? ml = 5 X 0.1 = 0.5 ml 10 X 0.1 = 1.0 ml 15 X 0.1 = 1.5 ml 20 X 0.1 = 2.0 ml 25 X 0.1 = 2.5 ml Jar Testing Calculations

9 A jar test indicates that the required amount of phosphorus removal can be achieved using a dosage of 3 ml of a 1% ferric chloride solution in a liter of wastewater. What is the dosage in mg/L? C 1 X V 1 = C 2 X V 2 10,000 mg/L X 3 ml = ? mg/L X 1000 ml 10,000 mg/L X 3 ml 1000 ml = 30 mg/L

10 Jar Testing Using the 1000 ml graduated cylinder, add 1000 ml wastewater to each beaker. Using the graduated pipets, dose each beaker with the desired concentration of metal salt or polymer, increasing concentration from left to right. Operate stirrer to simulate plant process. Determine best dosage level by analysis of supernatant Procedure

11 Gang Stirrer Addition of Chemicals Rapid Mix Slow Mix - Flocculate Settle

12 Determine which dosage is best for meeting requirements

13 Chemical Handling After jar testing has been used to determine best chemical dosage in mg/L, pounds of chemical needed per day into a given flow or volume can be calculated. lbs/day = MGD X 8.34 lbs/gal X mg/L ie) A dosage of 25 mg/L Ferric Chloride is needed. The flow to be treated is 350,000 gallons per day. How many lbs of Ferric Chloride will have to be fed each day? lbs/D = 0.35 MGD X 8.34 lbs/gal X 25 mg/L = 73 lbs/D FeCL 3

14 Chemical Handling Given lbs/d of dry chemical to feed, need to calculate the gallons of solution to feed. Specific Gravity = the number of times heavier or lighter the solution is than water 1 gallon of water weighs 8.34 lbs Specific gravity of water = 1.000 Specific Gravity

15 If a solution has a Specific Gravity of 1.510, then this solution is 1.510 times heavier than water. 8.34 lbs/gal X 1.510 = 12.59 lbs/gal A solution with a Specific Gravity of 0.750 would weigh: 8.34 lbs/gal X 0.750 = 6.255 lbs/gal Specific Gravity

16 Obtain S.G. from Supplier or by Using a Hydrometer

17 Chemical Handling Strength of Concentrated Solutions are usually given as Percent by Weight In 100 lbs of a 35 % solution there are 35 lbs of dry chemical 35 % = 35 lbs dry chem 100 lbs solution 35 % = 35 per 100

18 If the weight of a gallon of solution is known, the weight of dry chemical in each gallon of the solution can be calculated: A 40% solution has a specific gravity of 1.43 A. Determine the weight of a gallon of the solution. 8.34 lbs/gallon X 1.43 = 11.93 lbs per gallon liquid B. Determine the lbs of dry chemical in each gallon. 11.93 lbs X= 4.77 lbs dry chem/ gal 40 lbs dry chem 100 lbs solution OR 11.93 lbs X = 4.77 lbs dry chem/ gal0.40

19 Chemical Handling Feed Rates If the lbs per day of chemical to be fed is known, and we know the pounds of dry chemical in each gallon of the solution, the gallons per day of solution to be fed can be calculated. lbs/day dry chem needed lbs dry chem / gal = gal/day soln to be fed

20 ie) 150 lbs per day ferric chloride are to be fed. The solution to be used is 38% with a specific gravity of 1.413. Calculate the gallons of solution to feed each day. 8.34 lbs/gal X 1.413 = 11.78 lbs/gal 0.38 X 11.78 lbs/gal = 4.48 lbs dry chem/gal 150 lbs/d dry chem needed 4.48 lbs dry chem/gal = 33.5 gal/day Chemical Handling

21 Pumping Rate Calculations Given gallons per day of chemical needed, calculate gallons per minute Gal Day X 1 Day 24 Hrs X 1 Hr 60 Min = Gal Min

22 mls Sec X 1 Gal 3785 mls X 60 Sec Min = Gal Min Calibration of Chemical Feed Pump 1. Set pump at known setting 2. Start pump and collect at the application point a measured amount in a known period of time. 3. Repeat for various settings. 4. Convert to needed units.

23 Chemical Feed Summary 1. Determine lbs/day of dry chemical needed. 2. Determine the weight (lbs/gal) of the soln. 3. Determine weight of dry chem in each gallon. 4. Determine gallons per day to be fed. 5. To Determine pumping rate in gallons/minute: gal Day X 1 Day 24 Hrs X 1 Hr 60 Min = Gal Min mg/L X MGD X 8.34 lbs/gal = lbs/d dry chem 8.34 lbs/gal X Sp.Gr. = lbs/gal (liquid weight) lbs/gal (liquid wt) X % soln = lbs dry/gallon Lbs dry needed per day = gal/day to be fed lbs dry/gallon

24 Chemical Handling Calculations Example Problem #1 1. Calculate the pounds per day of FeCl 3 that must be added to the wastewater flow. Jar Test Results23 mg/L FeCl 3 Dosage Needed Wastewater Flow300,000 gallons per day Ferric Chloride Solution Specific Gravity1.500 Concentration45 % Work Calculations on Separate Paper Answers Given on Next Slides 2. Calculate the weight in pounds of a gallon of the FeCl 3 solution. 3. Calculate the pounds of dry FeCl 3 per gallon of the solution. 4. Calculate the number of gallons of the FeCl 3 solution that must be fed per day. 5. How many gallons per minute must the chemical feed pump be set to deliver ?

25 Chemical Handling Calculations Example Problem #1 1. Calculate the pounds per day of FeCl 3 that must be added to the wastewater flow. Jar Test Results23 mg/L FeCl 3 Dosage Needed Wastewater Flow300,000 gallons per day Ferric Chloride Solution Specific Gravity1.500 Concentration45 % lbs/d = 23 mg/L X 8.34 lbs/gal X 0.30 MGD = 57.55 lbs/d

26 3. Calculate the pounds of dry FeCl 3 per gallon of the solution. 12.51 lbs/gal X 0.45 = 5.63 lbs dry/gallon 8.34 lbs/gal X 1.500 = 12.51 lbs/gal 2. Calculate the weight in pounds of a gallon of the FeCl 3 solution.

27 4. Calculate the number of gallons of the FeCl 3 solution that must be fed per day. 5. How many gallons per minute must the chemical feed pump be set to deliver ? 57.55 lbs/day needed 5.63 lbs dry/gallon = 10.2 gallons/day 10.2 gal Day X 1 Day 24 Hrs X 1 Hr 60 Min = 0.0071 Gal Min 0.0071 gal min X 3785 mls gal 26.12 ml/min=

28 Calculate the number of milliliters per minute that must be fed of a 49 % Aluminum Sulfate solution with a specific gravity of 1.33 to dose a flow of 150,000 gallons per day at 25 mg/L. Chemical Handling Calculations Example Problem #2 Work Calculations on Separate Paper Answers Given on Next Slide

29 Calculate the number of milliliters per minute that must be fed of a 49 % Aluminum Sulfate solution with a specific gravity of 1.33 to dose a flow of 150,000 gallons per day at 25 mg/L. 25 mg/L X 0.15 MGD X 8.34 lbs/gal = 31.28 lbs/d 8.34 lbs/gal X 1.33 X 0.49 = 5.44 lbs dry/gal 31.28 lbs dry needed/day 5.44 lbs dry/gallon = 5.75 gal/day 15.11 ml/min 5.75 gal day X 3785 mls gal 1 day 1440 min X = Chemical Handling Calculations Example Problem #2

30 120 gallons/day of a ferric chloride solution was fed into a flow of 2.5 MGD. The ferric chloride solution had a concentration of 40 % and a specific gravity of 1.430. Calculate the chemical dosage in mg/L. Chemical Handling Calculations Example Problem #3 Work Calculations on Separate Paper Answers Given on Next Slide

31 120 gallons/day of a ferric chloride solution was fed into a flow of 2.5 MGD. The ferric chloride solution had a concentration of 40 % and a specific gravity of 1.430. Calculate the chemical dosage in mg/L. 8.34 lbs/gal X 1.430 X 0.40 = 4.77 lbs dry/gal 120 gal X 4.77 lbs dry/gal = 572.5 lbs dry/day 572 lbs/day = 2.5 MGD X 8.34 lbs/gal X mg/L 572 lbs/day 2.5 MGD X 8.34 lbs/gal = 27.5 mg/L Chemical Handling Calculations Example Problem #3

32 Jar Testing of Chemical Dosages Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit


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