1. Well Disinfection Math for Water Technology MTH 082 Lecture
Chapter 8 Water Sources and Storage (Price).
Well Casing Disinfection (Pg )
Oregon Department of Human Services
Coliform Bacteria and Well Disinfection
Disinfection Section AWWA
Oregon State University. How To Disinfect A Well and Water System

2. Objectives Well Drawdown Well disinfection Well yield
Reading assignment:
Chapter 8 Water Sources and Storage (Price).
Well Casing Disinfection (Pg )
Oregon Department of Human Services
Coliform Bacteria and Well Disinfection
Disinfection Section 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
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
Given
Formula
Solve:
307 ft
-27 ft
27 ft
0 ft

5. Fluid Pressure Practice Problem
A B C
Elevation at surface (m)
Depth of Piezometer (m)
Depth to water
(m below surface)
145
148
165
Hydraulic Head
Pressure Head
Elevation Head 70 23 15 75
125
150
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, 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*****50 mg/L****** EXISTING WELLS*****100 mg/L
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
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?
Given
Formula
Solve:
total = 740 gal, time = 5 minutes
Well yield = Flow gallons
Duration of Test, min
Well yield = gallons = 148 gpm
5 min
67 gpm
148 gpm
3700 gpm
0 gpm

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?
Given
Formula
Solve:
Cl= 65/100 D=18 in=1.5 ft Well =120 ft
220 ft 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)= gal
(50 mg/L)( MG)(8.34 lb/gal) = 1.01lbs
65/100
2 lbs
1 lbs
.02 lbs
0.65 lbs

15. Lbs/day= (mg/L)(MG)(8.34 lb/gal)
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 = MG
Lbs/day= (mg/L)(MG)(8.34 lb/gal)
0.5 Lbs/day= (X mg/L)( MG)(8.34 lb/gal)
X= 0.5/( MG)(8.34)
X= 109 mg/L
Given
Formula
Solve:
1209 mg/L
109 mg/L
1 mg/L
0. 5 mg/L

16. A new well is to be disinfected with chlorine at a dosage of 50 mg/L
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?
Given
Formula
Solve:
D=6 in=0.5 ft Well =120 ft
220 ft 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)= gal
(50 mg/L)( MG)(8.34 lb/gal) = 0.07
2 lbs
0.07 lbs
5.70 lbs
1.70 lbs

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