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Well Disinfection

Well Disinfection. Math for Water Technology MTH 082 Lecture Well Disinfection 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

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Well Disinfection

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  1. Well Disinfection Math for Water Technology MTH 082 Lecture Well Disinfection 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

  2. Objectives 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 • Well Drawdown • Well disinfection • Well yield

  3. Depth of Piezometer Depth to Water H2O Table (hp) Pressure Head: pressure (density )(gravity) Total Head (h): (z) + pressure (density )(gravity) Z = elevation of base of piezometer above or below some datum (Sea Level)

  4. 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? • 307 ft • -27 ft • 27 ft • 0 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 Given Formula Solve:

  5. Fluid Pressure Practice Problem 145 148 165 70 23 15 75 125 150 A B C Elevation at surface (m) 225 225 225 Depth of Piezometer (m) 150 100 75 Depth to water 80 77 60 (m below surface) Hydraulic Head Pressure Head Elevation Head What is hydraulic head at A, B, C?=elev. of H20 in piezometer What is pressure head at A, B, C?=height of H20 above piezometer What is elevation head at A, B, C?=height @ BOTTOM of piezometer (ABOVE/BELOW REFERENCE)

  6. 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 Cl2)(water in well casing MG)(8.34 lb/gal) Chlorine lbs = chlorine lbs % available chlorine 100

  7. Well Problems • New Wells Well casing disinfection lbs= (dose mg/L Cl2)(water in well casing MG)(8.34 lb/gal) NEW WELLS*****50 mg/L****** EXISTING WELLS*****100 mg/L Chlorine lbs = chlorine lbs % available chlorine 100

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

  9. 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 • Disinfect the well again without skipping any steps. • Place the well into service anyway. It will be fine after a couple of hours. • Place the well into service, but maintain twice the normal residual chlorine concentration for a few days. • Abandon the well. If it isn't clean now, it never will be.

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

  11. What concentration of residual chlorine should be maintained for 24 hours in a newly constructed well? • 50 mg/L • 50 ug/L • 25 mg/L • 25 ug/L

  12. After a routine repair to an existing well, how much chlorine residual is required in the well to ensure adequate disinfection? • 100 mg/L • 200 mg/L • 400 mg/L • 1000 mg/L

  13. 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? • 67 gpm • 148 gpm • 3700 gpm • 0 gpm Given Formula Solve: total = 740 gal, time = 5 minutes Well yield = Flow gallons Duration of Test, min Well yield = 740 gallons = 148 gpm 5 min

  14. 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? • 2 lbs • 1 lbs • .02 lbs • 0.65 lbs Given Formula Solve: 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)(D2)(H) = ft3 (0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft3)= 1585 gal (50 mg/L)(.001585 MG)(8.34 lb/gal) = 1.01lbs 65/100

  15. 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? • 1209 mg/L • 109 mg/L • 1 mg/L • 0. 5 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 Given Formula Solve:

  16. 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? • 2 lbs • 0.07 lbs • 5.70 lbs • 1.70 lbs Given Formula Solve: D=6 in=0.5 ft Well =120 ft 220 ft - 100 ft = 120 ft water in well (0.785)(D2)(H) = ft3 (mg/L)(MG)(8.34 lb/gal) (0.785)(0.5 ft)(0.5 ft) (120 ft)(7.48 gal/ft3)= 176 gal (50 mg/L)(.000176 MG)(8.34 lb/gal)= 0.07

  17. Today’s objective: Well Disinfection, Well Casing, Well yield and chlorine dosage of new wells been met? • Strongly Agree • Agree • Neutral • Disagree • Strongly Disagree

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