2. Groundwater Math Math for Water Technology MTH 082 Lecture 2 Well Drawdown, Well yield, Specific Capacity, Well Disinfection Course Reader Week 2 Chapter 8 Water Sources and Storage (Price). Well Casing Disinfection (Pg 174-177) Oregon Department of Human Services Coliform Bacteria and Well Disinfection Disinfection Section 02675. AWWA Oregon State University. How To Disinfect A Well and Water System

3. Objectives 1.Well Drawdown (static and pumping water levels) 2.Well Yield (drawdown test) 3.Specific Capacity 4.Well disinfection 1.Well Drawdown (static and pumping water levels) 2.Well Yield (drawdown test) 3.Specific Capacity 4.Well disinfection Reading assignment: Chapter 8 Water Sources and Storage (Price). Well Casing Disinfection (Pg 174-177) Oregon Department of Human Services Coliform Bacteria and Well Disinfection Disinfection Section 02675. AWWA Oregon State University. How To Disinfect A Well and Water System Reading assignment: Chapter 8 Water Sources and Storage (Price). Well Casing Disinfection (Pg 174-177) Oregon Department of Human Services Coliform Bacteria and Well Disinfection Disinfection Section 02675. AWWA Oregon State University. How To Disinfect A Well and Water System

4. Well Problems Drawdown ft = pumping water level – static water level ft Well yield = Flow gallons duration of Test, min -small wells = (gpm, gph) Large wells= (ft 3 /sec or m 3 /sec) Specific capacity, 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 -small wells = (gpm, gph) Large wells= (ft 3 /sec or m 3 /sec) Specific capacity, 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

5. Static H 2 O Table Well screen Gravel Pack Grout Seal minimum 10 ft Drawdown Pumping water level Cone of Depression Well casing Submersible Well pump Well slab Sanitary Seal Pump House 1. Electricity 2. Disinfection

6. Given Formula Solve: Given Formula Solve: The static water level for a well is 89 ft. If the pumping water level is 97 ft, what is the well drawdown?? Static WL= 89 ft, Pumped WL=97 ft Drawdown ft = pumping water level – static water level ft Drawdown = 97 ft- 89 ft Drawdown = 8 ft Static WL= 89 ft, Pumped WL=97 ft Drawdown ft = pumping water level – static water level ft Drawdown = 97 ft- 89 ft Drawdown = 8 ft 1.186 ft 2.-8 ft 3.8 ft 4.97 ft 1.186 ft 2.-8 ft 3.8 ft 4.97 ft

7. 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

8. Given Formula Solve: Given Formula Solve: The static water level for a well is 105 ft. If the pumping water level is 128 ft, what is the well drawdown?? Static WL= 105 ft, Pumped WL=105 ft Drawdown ft = pumping water level – static water level ft Drawdown = 128 ft- 105 ft Drawdown = 23 ft Static WL= 105 ft, Pumped WL=105 ft Drawdown ft = pumping water level – static water level ft Drawdown = 128 ft- 105 ft Drawdown = 23 ft 1.133 ft 2.128 ft 3.23 ft 4.105 ft 1.133 ft 2.128 ft 3.23 ft 4.105 ft

9. Given Formula Solve: Given Formula Solve: A sounding line is used to determine the static water level in a well. The air pressure applied is 4.5 psi and the length of the sounding line is 150 ft. If the pumping water level is 168 ft, what is the drawdown? Static WL= ? ft, Pumped WL= ft 4.5 psi (2.31 ft/1 psi) = 10.4 ft WL in well ft = length of sounding line,ft– Water depth in sounding line, ft WL in well ft =150 ft- 10.4 ft WL in well ft = 139 ft Drawdown = 168 ft-139.6 ft = 28.4 ft Static WL= ? ft, Pumped WL= ft 4.5 psi (2.31 ft/1 psi) = 10.4 ft WL in well ft = length of sounding line,ft– Water depth in sounding line, ft WL in well ft =150 ft- 10.4 ft WL in well ft = 139 ft Drawdown = 168 ft-139.6 ft = 28.4 ft 1.168 ft 2.28.4 ft 3.139 ft 4.18 ft 1.168 ft 2.28.4 ft 3.139 ft 4.18 ft

10. H 2 O Table Depth to Water (h p ) Pressure Head: pressure (density  )(gravity) (h p ) Pressure Head: pressure (density  )(gravity) Total Head (h): (z) + pressure (density  )(gravity) Total Head (h): (z) + pressure (density  )(gravity) Z = elevation of base of piezometer above or below some datum (Sea Level) Depth of Piezometer Depth of Piezometer

11. Fluid Pressure Practice Problem ABCABC ABCABC Elevation at surface (m)225225225 Depth of Piezometer (m)15010075 Depth to water807760 (m below surface) Elevation at surface (m)225225225 Depth of Piezometer (m)15010075 Depth to water807760 (m below surface) What is hydraulic head at A, B, C?=elev. of H 2 0 in piezometer What is pressure head at A, B, C?=height of H 2 0 above piezometer What is elevation head at A, B, C?=height @ BOTTOM of piezometer (ABOVE/BELOW REFERENCE) What is hydraulic head at A, B, C?=elev. of H 2 0 in piezometer What is pressure head at A, B, C?=height of H 2 0 above piezometer What is elevation head at A, B, C?=height @ BOTTOM of piezometer (ABOVE/BELOW REFERENCE) Hydraulic Head Pressure Head Elevation Head Hydraulic Head Pressure Head Elevation Head 145 148165 70 2315 75 125150

12. Well Yield Well Yield: Is the rate of water withdrawal that a well can supply over a long period of time. Well yield = Flow gallons duration of Test, min Small wells = (gpm, gph) Large wells= (ft 3 /sec or m 3 /sec) In the U.S. most building or health departments who must approve a private well water supply when issuing a final certificate of occupancy for new construction want to see 3 to *5* gallons per minute or 3-5 gpm. Well Yield: Is the rate of water withdrawal that a well can supply over a long period of time. Well yield = Flow gallons duration of Test, min Small wells = (gpm, gph) Large wells= (ft 3 /sec or m 3 /sec) In the U.S. most building or health departments who must approve a private well water supply when issuing a final certificate of occupancy for new construction want to see 3 to *5* gallons per minute or 3-5 gpm.

13. 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

14. Given Formula Solve: Given Formula Solve: During a test for well yield, a total of 835 gallons were pumped from the well. If the well yield test lasted 5 minutes, what was the well yield in gph? total = 835 gal, time = 5 minutes Well yield = Flow gallons Duration of Test, min Well yield =835 gallons = 167 gpm (60 min/1 hr) = 10,020 gph 5 min total = 835 gal, time = 5 minutes Well yield = Flow gallons Duration of Test, min Well yield =835 gallons = 167 gpm (60 min/1 hr) = 10,020 gph 5 min 1.148 gph 2.835 gph 3.4175 gpm 4.10,020 gph 1.148 gph 2.835 gph 3.4175 gpm 4.10,020 gph

15. Specific Capacity Specific Capacity: Is the rate of well yield per unit of drawdown. Specific capacity, gpm/ft = (Well yield gpm) (Drawdown ft) Specific Capacity: Is the rate of well yield per unit of drawdown. Specific capacity, gpm/ft = (Well yield gpm) (Drawdown ft)

16. Given Formula Solve: Given Formula Solve: It was determined that a well had a yield of 200 gpm after 25 ft of drawdown, What is the wells specific capacity? Well yield = 200 gpm Drawdown=25 ft Specific Capacity = Well Yield (gpm) Drawdown (ft) SC = 200 gpm = 8 gpm per 1 ft drawdown 25 ft Well yield = 200 gpm Drawdown=25 ft Specific Capacity = Well Yield (gpm) Drawdown (ft) SC = 200 gpm = 8 gpm per 1 ft drawdown 25 ft 1.15 gpm per 1 ft of drawdown 2.25 gpm per 1 ft of drawdown 3.200 gpm per 1 ft of drawdown 4.8 gpm per 1 ft of drawdown 1.15 gpm per 1 ft of drawdown 2.25 gpm per 1 ft of drawdown 3.200 gpm per 1 ft of drawdown 4.8 gpm per 1 ft of drawdown

17. Given Formula Solve: Given Formula Solve: A well produces 1000 gpm, If the drawdown was 38.2 ft, what is the wells specific capacity in gpm/ft? Well yield = 1000 gpm Drawdown=38.2 ft Specific Capacity = Well Yield (gpm) Drawdown (ft) SC = 1000 gpm = 26.2 gpm/ft 38.2 ft Well yield = 1000 gpm Drawdown=38.2 ft Specific Capacity = Well Yield (gpm) Drawdown (ft) SC = 1000 gpm = 26.2 gpm/ft 38.2 ft 1.1000 gpm/ft 2.38.2 gpm/ft 3.26.2 gpm/ft 4.3820 gpm/ft 1.1000 gpm/ft 2.38.2 gpm/ft 3.26.2 gpm/ft 4.3820 gpm/ft

18. Well Disinfection New Wells: Well casing disinfection lbs= (dose mg/L Cl 2 )(water in well casing MG)(8.34 lb/gal) Well casing Volume:( 0.785)(Diameter)(Diamter)(length) Volume: ft 3 to gallons (7.48 gal/1 ft 3 ) Volume: (1 gal) (1MG/1,000,000 gal) NEW WELLS*****50 mg/L****** EXISTING WELLS*****100 mg/L Chlorine lbs = chlorine lbs % available chlorine 100 New Wells: Well casing disinfection lbs= (dose mg/L Cl 2 )(water in well casing MG)(8.34 lb/gal) Well casing Volume:( 0.785)(Diameter)(Diamter)(length) Volume: ft 3 to gallons (7.48 gal/1 ft 3 ) Volume: (1 gal) (1MG/1,000,000 gal) NEW WELLS*****50 mg/L****** EXISTING WELLS*****100 mg/L Chlorine lbs = chlorine lbs % available chlorine 100

19. What is the purpose of surging? 1.To clean mineral deposits from well screens. 2.To remove blockages from the distribution system. 3.To backwash filters rapidly. 4.To prepare pump motors for erratic power supplies. 1.To clean mineral deposits from well screens. 2.To remove blockages from the distribution system. 3.To backwash filters rapidly. 4.To prepare pump motors for erratic power supplies.

20. A new well has been disinfected according to the standard operating procedure, but a fecal coliform sample taken after disinfection still shows colony growth. The operator should 1.Disinfect the well again without skipping any steps. 2.Place the well into service anyway. It will be fine after a couple of hours. 3.Place the well into service, but maintain twice the normal residual chlorine concentration for a few days. 4.Abandon the well. If it isn't clean now, it never will be. 1.Disinfect the well again without skipping any steps. 2.Place the well into service anyway. It will be fine after a couple of hours. 3.Place the well into service, but maintain twice the normal residual chlorine concentration for a few days. 4.Abandon the well. If it isn't clean now, it never will be.

21. Chlorine is an effective treatment for well screens. It helps to remove this material. 1.Slime from iron-oxidizing bacteria 2.Biofilms from ammonia- oxidizing bacteria 3.Iron and manganese oxides 4.Calcium carbonate deposits 1.Slime from iron-oxidizing bacteria 2.Biofilms from ammonia- oxidizing bacteria 3.Iron and manganese oxides 4.Calcium carbonate deposits

22. What concentration of residual chlorine should be maintained for 24 hours in a newly constructed well? 1.50 mg/L 2.50 ug/L 3.25 mg/L 4.25 ug/L 1.50 mg/L 2.50 ug/L 3.25 mg/L 4.25 ug/L

23. After a routine repair to an existing well, how much chlorine residual is required in the well to ensure adequate disinfection? 1.100 mg/L 2.200 mg/L 3.400 mg/L 4.1000 mg/L 1.100 mg/L 2.200 mg/L 3.400 mg/L 4.1000 mg/L

24. Given Formula Solve: Given Formula Solve: A well casing contains 550 gal of water. If 0.5 lbs of chlorine were used in the disinfection, what was the chlorine dosage in mg/L? 550 gal = 0.000550 MG Lbs/day= (mg/L)(MG)(8.34 lb/gal) 0.5 Lbs/day= (X mg/L)(.000550 MG)(8.34 lb/gal) X= 0.5/(0.00050 MG)(8.34) X= 109 mg/L 1.1209 mg/L 2.109 mg/L 3.1 mg/L 4.0. 5 mg/L 1.1209 mg/L 2.109 mg/L 3.1 mg/L 4.0. 5 mg/L

25. MINDBENDERS! Proceed at your Own RISK

26. Given Formula Solve: Given Formula Solve: A new well is to be disinfected with chlorine at a dosage of 50 mg/L. If the well casing diameter is 6 inches and the length of the water filled casing is 120 ft, how many lbs of chlorine will be required? D=6 in=0.5 ft Well =120 ft 220 ft - 100 ft = 120 ft water in well (0.785)(D 2 )(H) = ft 3 (mg/L)(MG)(8.34 lb/gal) (0.785)(0.5 ft)(0.5 ft) (120 ft)(7.48 gal/ft 3 )= 176 gal (50 mg/L)(.000176 MG)(8.34 lb/gal)= 0.07 D=6 in=0.5 ft Well =120 ft 220 ft - 100 ft = 120 ft water in well (0.785)(D 2 )(H) = ft 3 (mg/L)(MG)(8.34 lb/gal) (0.785)(0.5 ft)(0.5 ft) (120 ft)(7.48 gal/ft 3 )= 176 gal (50 mg/L)(.000176 MG)(8.34 lb/gal)= 0.07 1.2 lbs 2.0.07 lbs 3.5.70 lbs 4.1.70 lbs 1.2 lbs 2.0.07 lbs 3.5.70 lbs 4.1.70 lbs

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: To practice math associated with Well Drawdown, Well yield, Specific Capacity, and Well Disinfection has 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