well disinfection math for water technology mth 082 lecture well disinfection chapter 8 water...
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Well DisinfectionWell Disinfection
Math for Water TechnologyMTH 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
Math for Water TechnologyMTH 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
ObjectivesObjectives
1. Well Drawdown
2. Well disinfection
3. Well yield
1. Well Drawdown
2. Well disinfection
3. Well yield
Reading assignment: Chapter 8 Water Sources and Storage (Price). Well Casing Disinfection (Pg 174-177)Oregon Department of Human Services Coliform Bacteria and Well DisinfectionDisinfection 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 DisinfectionDisinfection Section 02675. AWWA Oregon State University. How To Disinfect A Well and Water System
H2O TableH2O TableDepth to WaterDepth to Water
(hp) Pressure Head: pressure (density )(gravity)
(hp) 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)
Z = elevation of base of piezometer above or below some datum (Sea Level)
Depth of Piezometer Depth of Piezometer
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?
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 2
7 ft
0 ft
0% 0%
83%
17%
Static WL= 140 ft, Pumped WL=167 ft
Drawdown ft = pumping water level – static water level ft
Drawdown = 167 ft- 140 ftDrawdown = 27 ft
Static WL= 140 ft, Pumped WL=167 ft
Drawdown ft = pumping water level – static water level ft
Drawdown = 167 ft- 140 ftDrawdown = 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
Fluid Pressure Practice ProblemFluid Pressure Practice ProblemA B CA B C
Elevation at surface (m) 225 225 225Depth of Piezometer (m) 150 100 75Depth to water 80 77 60(m below surface)
Elevation at surface (m) 225 225 225Depth of Piezometer (m) 150 100 75Depth to water 80 77 60(m below surface)
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)
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)
Hydraulic HeadPressure HeadElevation Head
Hydraulic HeadPressure HeadElevation Head
145 148 16570 23 1575 125 150
Well ProblemsWell 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
• 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
Well ProblemsWell 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
• 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
What is the purpose of surging?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.
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
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
Dis
infe
ct th
e ...
Pla
ce th
e wel
l...
Pla
ce th
e wel
l...
Aban
don the
we.
..
25% 25%25%25%
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.
Chlorine is an effective treatment for well screens. It helps to
remove this material.
Chlorine is an effective treatment for well screens. It helps to
remove this material.
Slim
e fro
m ir
o...
Bio
film
s fro
m ..
.
Iron a
nd m
anga.
..
Cal
cium
car
bon...
25% 25%25%25%
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
What concentration of residual chlorine should be maintained
for 24 hours in a newly constructed well?
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
25% 25%25%25%
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
After a routine repair to an existing well, how much chlorine residual is required in the well to ensure adequate disinfection?
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
100
0 m
g/L
25% 25%25%25%
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
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?
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
370
0 gpm
0 g
pm
8%0%0%
92%
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
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?
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 lb
s
1 lb
s
.02
lbs
0.6
5 lb
s
7%14%
0%
79%
Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft220 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
Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft220 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
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
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?
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?
120
9 m
g/L
109
mg/L
1 m
g/L
0. 5
mg/L
25% 25%25%25%
550 gal = 0.000550 MGLbs/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
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?
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 lb
s
0.0
7 lb
s
5.7
0 lb
s
1.7
0 lb
s
25% 25%25%25%
D=6 in=0.5 ft Well =120 ft220 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
D=6 in=0.5 ft Well =120 ft220 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
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
Today’s objective: Well Disinfection, Well Casing, Well yield and chlorine dosage of
new wells been met?
Today’s objective: Well Disinfection, Well Casing, Well yield and chlorine dosage of
new wells been met?
Stro
ngly A
gree
Agre
e
Neu
tral
Dis
agre
e
Stro
ngly D
isag
ree
86%
14%
0%0%0%
1. Strongly Agree
2. Agree
3. Neutral
4. Disagree
5. Strongly Disagree
1. Strongly Agree
2. Agree
3. Neutral
4. Disagree
5. Strongly Disagree