1. Filter Loading and Backwash Rates Well Yields and Chlorine Dosage in Waterworks Operation Math for Water Technology MTH 082 Lecture Chapter 4 & 8- Applied Math for Water Plant Operators Loading Rate Calculations (pg 62-66); Well yield (163-177) Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203)

2. Objectives 1.Filter overviews 2.Filter loading and backwash rates 3.Well yield 4.Chlorine dosage for a new well 1.Filter overviews 2.Filter loading and backwash rates 3.Well yield 4.Chlorine dosage for a new well Reading assignment: Chapter 4 & 8- Applied Math for Water Plant Operators Loading Rate Calculations (pg 62-66); Well yield (163-177) Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203) Reading assignment: Chapter 4 & 8- Applied Math for Water Plant Operators Loading Rate Calculations (pg 62-66); Well yield (163-177) Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203)

3. Conventional Treatment Conventional Treatment – common treatment steps used to remove turbidity from the initial source water. 2. Flocculation 3. Sedimentation 4. Filtration 1. Coagulation Pretreatment Raw Water Rapid or flash Mixing Alum, polymer Rapid or flash Mixing Alum, polymer Slow Fast Sludge Washwater 5. Clear Well Disinfection 5. Clear Well Disinfection Distri bution Distri bution Clearwell Backwash pumps Clearwell Backwash pumps Chlorination Ozone UV Chlorination Ozone UV

4. High Rate Filter Dual or Multi Media Filter -4 times faster then rapid sand -granular activated carbon -anthracite coal -garnet sand and gravel -backwash ready (70 hr longevity) -excellent water quality Dual or Multi Media Filter -4 times faster then rapid sand -granular activated carbon -anthracite coal -garnet sand and gravel -backwash ready (70 hr longevity) -excellent water quality

5. High Rate Filters Anthracite Coal Fine sand Garnet sand Garnet sand

6. High Rate Filter Dualmedia Filter Anthracite Coal Fine sand Garnet sand Garnet sand Monomedia Coarse sand Monomedia Coarse sand Monomedia Filter

7. Dual-media and multimedia filters 1.Require an extremely deep bed 2.Can operate at three or four times the rate of sand filters 3.Cannot reduce turbidity 4.Do not require backwashing 1.Require an extremely deep bed 2.Can operate at three or four times the rate of sand filters 3.Cannot reduce turbidity 4.Do not require backwashing

8. When mixed media filters composed of garnet, sand, and crushed anthracite coal are used, which of the following describes their placement in the filter bed? 1.Anthracite coal on top, garnet in the middle, and sand on the bottom 2.Garnet on top, anthracite coal in the middle, and sand on the bottom 3.Sand on top, anthracite coal in the middle, and garnet on the bottom 4.Anthracite coal on top, sand in the middle, and garnet on the bottom 1.Anthracite coal on top, garnet in the middle, and sand on the bottom 2.Garnet on top, anthracite coal in the middle, and sand on the bottom 3.Sand on top, anthracite coal in the middle, and garnet on the bottom 4.Anthracite coal on top, sand in the middle, and garnet on the bottom

9. In a filter using gravel, anthracite, and sand, the anthracite should be? 1.The top layer of media 2.Beneath the gravel 3.Between the sand and the gravel 4.Mixed with the sand 1.The top layer of media 2.Beneath the gravel 3.Between the sand and the gravel 4.Mixed with the sand

10. Pilot Filters Anthracite Coal Fine sand Garnet sand

11. The main action of a mixed media filter is: 1.Straining 2.Disinfecting 3.Coagulating 4.None of the above 1.Straining 2.Disinfecting 3.Coagulating 4.None of the above

12. Backwash Suspended particles entrapped by filter media. Accumulation occurs: –head loss within the filter to reach excessively high levels (6 to 8 feet of hydraulic head). -Particles pass through the filter, water turbidities reach unacceptable levels -Rule Backwash at 0.1 NTU -SWTR Allows 0.3 NTU. Suspended particles entrapped by filter media. Accumulation occurs: –head loss within the filter to reach excessively high levels (6 to 8 feet of hydraulic head). -Particles pass through the filter, water turbidities reach unacceptable levels -Rule Backwash at 0.1 NTU -SWTR Allows 0.3 NTU.

13. Head Loss Clean filter =0 psi= one foot of head loss on a new filter As filter clogs more negative pressure Pressure builds in a linear fashion -2.5 to -4.0 psi = 6 to 10 ft of head loss More clogged greater the head loss Remember 1 ft of water column = 0.433 psi 2.31 ft of water for 1 psi change Clean filter =0 psi= one foot of head loss on a new filter As filter clogs more negative pressure Pressure builds in a linear fashion -2.5 to -4.0 psi = 6 to 10 ft of head loss More clogged greater the head loss Remember 1 ft of water column = 0.433 psi 2.31 ft of water for 1 psi change

14. Filter Backwash Some plants use head loss, some use time Some plants use operator knowledge and turbidity Each operator has their own scheme!! Parents and diapers-- Some plants use head loss, some use time Some plants use operator knowledge and turbidity Each operator has their own scheme!! Parents and diapers--

15. The most critical criterion for determining when a mixed media filter should be backwashed is: 1.Filter effluent quality 2.Flow rate 3.Head loss 4.Visual inspection of the filter surface 1.Filter effluent quality 2.Flow rate 3.Head loss 4.Visual inspection of the filter surface

16. Filtration Rate Calculation Unit Filter Run= (Total gal filtered gal) Filter Area (sq ft) Filtration Rate= (flow gpm) Area (sq ft) Units will be gpm ft 2 ! Backwash Rate = (flow gpm) Area (sq ft) Units will be gpm ft 2 ! Unit Filter Run= (Total gal filtered gal) Filter Area (sq ft) Filtration Rate= (flow gpm) Area (sq ft) Units will be gpm ft 2 ! Backwash Rate = (flow gpm) Area (sq ft) Units will be gpm ft 2 ! Downward Upward

17. The total water filtered during a filter run (between backwashes) is 2,950,000 gal. If the filter is 15 ft by 20 ft, What is the unit filter run volume (UFRV)? L= 15 ft, W=20 ft; Rate 2,950,000 gal A=L X W UFVR= (Total gallons filtered g) Area (sq ft) A= 20 ft X 15 ft = 300 ft 2 UFVR= (2,950,000 gal) 300 (sq ft) UFVR = 9833 g/ft 2 L= 15 ft, W=20 ft; Rate 2,950,000 gal A=L X W UFVR= (Total gallons filtered g) Area (sq ft) A= 20 ft X 15 ft = 300 ft 2 UFVR= (2,950,000 gal) 300 (sq ft) UFVR = 9833 g/ft 2 Given Formula Solve: Given Formula Solve: 1. 9.833 g/ft 2 2.101.7 g/ft 2 3.9833 g/ft 2 4. 0.9833 g/ft 2 1. 9.833 g/ft 2 2.101.7 g/ft 2 3.9833 g/ft 2 4. 0.9833 g/ft 2

18. The total water filtered during a filter run (between backwashes) is 4.8 MG. If the filter is 20 ft by 30 ft, What is the unit filter run volume (UFRV)? L= 20 ft, W=30 ft; Rate 4.8 MG A=L X W UFVR= (Total gallons filtered g) Area (sq ft) A= 20 ft X 30 ft = 600 ft 2 UFVR= (4,800,000 gal) 600 (sq ft) UFVR = 8000 g/ft 2 L= 20 ft, W=30 ft; Rate 4.8 MG A=L X W UFVR= (Total gallons filtered g) Area (sq ft) A= 20 ft X 30 ft = 600 ft 2 UFVR= (4,800,000 gal) 600 (sq ft) UFVR = 8000 g/ft 2 Given Formula Solve: Given Formula Solve: 1. 96000 g/ft 2 2.8000 g/ft 2 3.800 g/ft 2 4. 0.008 g/ft 2 1. 96000 g/ft 2 2.8000 g/ft 2 3.800 g/ft 2 4. 0.008 g/ft 2

19. A filter 20 ft by 25 ft receives a flow of 1940 gpm. What is the filtration rate in gpm/ft 2 ? L= 20 ft, W=25 ft; Rate 1940 gpm A=L X W Filtration Rate= (flow gpm) Area (sq ft) A= 20 ft X 25 ft = 500 ft 2 Filtration Rate= (1940 gpm) 500 (sq ft) Filtration Rate = 3.9 gpm/ft 2 L= 20 ft, W=25 ft; Rate 1940 gpm A=L X W Filtration Rate= (flow gpm) Area (sq ft) A= 20 ft X 25 ft = 500 ft 2 Filtration Rate= (1940 gpm) 500 (sq ft) Filtration Rate = 3.9 gpm/ft 2 1. 3.9 gpd/ft 3 2.3.9 gpm/ft 2 3.0.25 gpm/ft 2 4. 0.25 gpd/ft 3 1. 3.9 gpd/ft 3 2.3.9 gpm/ft 2 3.0.25 gpm/ft 2 4. 0.25 gpd/ft 3 Given Formula Solve: Given Formula Solve:

20. A filter 20 ft by 35 ft receives a flow of 1530 gpm. What is the filtration rate in gpm/ft 2 ? L= 20 ft, W=35 ft; Rate 1530 gpm A=L X W Filtration Rate= (flow gpm) Area (sq ft) A= 20 ft X 35 ft = 700 ft 2 Filtration Rate= (1530 gpm) 700 (sq ft) Filtration Rate = 2.2 gpm/ft 2 L= 20 ft, W=35 ft; Rate 1530 gpm A=L X W Filtration Rate= (flow gpm) Area (sq ft) A= 20 ft X 35 ft = 700 ft 2 Filtration Rate= (1530 gpm) 700 (sq ft) Filtration Rate = 2.2 gpm/ft 2 1. 2.2 gpm/ft 2 2.2.2 gpm/ft 3 3.0.45 gpm/ft 2 4. 0.45 gpd/ft 3 1. 2.2 gpm/ft 2 2.2.2 gpm/ft 3 3.0.45 gpm/ft 2 4. 0.45 gpd/ft 3 Given Formula Solve: Given Formula Solve:

21. A filter 25 ft by 30 ft receives a flow of 3.3 MGD. What is the filtration rate in gpm/ft 2 ? L= 25 ft, W=30 ft;Rate 3.3 MG 1,000,000 gal 1 Day D 1MG 1440 min A=L X W Filtration Rate= (flow gpm) Area (sq ft) A= 25 ft X 30 ft = 750 ft 2 Filtration Rate= (2292 gpm) 750 (sq ft) Filtration Rate = 3.1 gpm/ft 2 L= 25 ft, W=30 ft;Rate 3.3 MG 1,000,000 gal 1 Day D 1MG 1440 min A=L X W Filtration Rate= (flow gpm) Area (sq ft) A= 25 ft X 30 ft = 750 ft 2 Filtration Rate= (2292 gpm) 750 (sq ft) Filtration Rate = 3.1 gpm/ft 2 1.3.1 gpm/ft 3 2.0.32 gpm/ft 2 3.3.1 gpm/ft 2 4.0.32 gpd/ft 3 1.3.1 gpm/ft 3 2.0.32 gpm/ft 2 3.3.1 gpm/ft 2 4.0.32 gpd/ft 3 Given Formula Solve: Given Formula Solve:

22. Given Formula Solve: Given Formula Solve: A filter 25 ft by 10 ft has a backwash rate of 3400 gpm. What is the filter backwash rate in gpm/ft 2 ? L= 25 ft, W=10 ft; Rate 3400 gpm A=L X W Backwash Rate = (flow gpm) Area (sq ft) A= 25 ft X 10 ft = 250 ft 2 Filter Backwash Rate= (3400 gpm) 250 (sq ft) Filter Backwash Rate = 13.6 gpm/ft 2 L= 25 ft, W=10 ft; Rate 3400 gpm A=L X W Backwash Rate = (flow gpm) Area (sq ft) A= 25 ft X 10 ft = 250 ft 2 Filter Backwash Rate= (3400 gpm) 250 (sq ft) Filter Backwash Rate = 13.6 gpm/ft 2 1.13.1 gpm/ft 3 2.0.074 gpm/ft 2 3.13.6 gpm/ft 2 4.136 gpd/ft 3 1.13.1 gpm/ft 3 2.0.074 gpm/ft 2 3.13.6 gpm/ft 2 4.136 gpd/ft 3

23. Given Formula Solve: Given Formula Solve: A filter 20 ft by 15 ft has a backwash rate of 4.5 MGD. What is the filter backwash rate in gpm/ft 2 ? L= 20 ft, W=15 ft; ft;Rate 4.5 MG 1,000,000 gal 1 Day D 1MG 1440min A=L X W Backwash Rate = (flow gpm) Area (sq ft) A= 20 ft X 15 ft = 300 ft 2 Filter Backwash Rate= (3125 gpm) 300 (sq ft) Filter Backwash Rate = 10.4 gpm/ft 2 L= 20 ft, W=15 ft; ft;Rate 4.5 MG 1,000,000 gal 1 Day D 1MG 1440min A=L X W Backwash Rate = (flow gpm) Area (sq ft) A= 20 ft X 15 ft = 300 ft 2 Filter Backwash Rate= (3125 gpm) 300 (sq ft) Filter Backwash Rate = 10.4 gpm/ft 2 1.10.4 gpm/ft 3 2.0.01 gpm/ft 2 3.10.4 gpm/ft 2 4.0.01 gpd/ft 3 1.10.4 gpm/ft 3 2.0.01 gpm/ft 2 3.10.4 gpm/ft 2 4.0.01 gpd/ft 3

24. Well Problems Drawdown ft = pumping water level – static water level ft Well yield = Flow gallons duration of Test, min Specific yield, gpm/ft = (Well yield gpm) (Drawdown ft) Well casing disinfection lbs= (dose mg/L Cl 2 )(water in well casing MG)(8.34 lb/gal) Chlorine lbs = chlorine lbs % available chlorine 100 Drawdown ft = pumping water level – static water level ft Well yield = Flow gallons duration of Test, min Specific yield, gpm/ft = (Well yield gpm) (Drawdown ft) Well casing disinfection lbs= (dose mg/L Cl 2 )(water in well casing MG)(8.34 lb/gal) Chlorine lbs = chlorine lbs % available chlorine 100

25. Given Formula Solve: Given Formula Solve: Before the pump is started the water level is measured at 140 ft. The pump is then started. If the pumping water level is determined to be 167 ft, what is the drawdown in ft? Static WL= 140 ft, Pumped WL=167 ft Drawdown ft = pumping water level – static water level ft Drawdown = 167 ft- 140 ft Drawdown = 27 ft Static WL= 140 ft, Pumped WL=167 ft Drawdown ft = pumping water level – static water level ft Drawdown = 167 ft- 140 ft Drawdown = 27 ft 1.307 ft 2.-27 ft 3.27 ft 4.0 ft 1.307 ft 2.-27 ft 3.27 ft 4.0 ft

26. Given Formula Solve: Given Formula Solve: During a five minute test for well yield, a total of 740 gallons are removed from the well. What is the well yield in gpm? total = 740 gal, time = 5 minutes Well yield = Flow gallons Duration of Test, min Well yield = 740 gallons = 148 gpm 5 min total = 740 gal, time = 5 minutes Well yield = Flow gallons Duration of Test, min Well yield = 740 gallons = 148 gpm 5 min 1.67 gpm 2.148 gpm 3.3700 gpm 4.0 gpm 1.67 gpm 2.148 gpm 3.3700 gpm 4.0 gpm

27. Given Formula Solve: Given Formula Solve: How many lbs of calcium hypochlorite (65% available chlorine) is required to disinfect a well if the casing is 18 inches in diameter and 220 ft long, with water level at 100 ft from the top of the well? The desired dose is 50 mg/L? Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft 220 ft - 100 ft = 120 ft water in well (0.785)(D 2 )(H) = ft 3 (0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft 3 )= 1585 gal (50 mg/L)(.001585 MG)(8.34 lb/gal) = 1.01lbs 65/100 Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft 220 ft - 100 ft = 120 ft water in well (0.785)(D 2 )(H) = ft 3 (0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft 3 )= 1585 gal (50 mg/L)(.001585 MG)(8.34 lb/gal) = 1.01lbs 65/100 1.2 lbs 2.1 lbs 3..02 lbs 4.0.65 lbs 1.2 lbs 2.1 lbs 3..02 lbs 4.0.65 lbs

28. Today’s objective: Filter Loading Rates, Filter Backwash Rates, Well yield and chlorine dosage of new wells been met? 1.Strongly Agree 2.Agree 3.Neutral 4.Disagree 5.Strongly Disagree 1.Strongly Agree 2.Agree 3.Neutral 4.Disagree 5.Strongly Disagree