Transcript Lecture Well drawdown and 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
Oregon State University. How To Disinfect A Well and Water
System
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
1. Well Drawdown
2. Well disinfection
3. Well yield
Depth of
Piezometer
Depth to Water
H2O Table
(hp) Pressure Head:
pressure
(density )(gravity)
Z = elevation of base of
piezometer above or below
some datum (Sea Level)
Total Head (h):
(z) + pressure
(density )(gravity)
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
Given
Formula
Drawdown ft = pumping water level – static water level ft
Solve:
Drawdown = 167 ft- 140 ft
83%
Drawdown = 27 ft
17%
ft
0
27
-2
7
ft
ft
0%
ft
0%
7
307 ft
-27 ft
27 ft
0 ft
30
1.
2.
3.
4.
Fluid Pressure Practice Problem
A
225
150
80
B
225
100
77
C
225
75
60
Elevation at surface (m)
Depth of Piezometer (m)
Depth to water
(m below surface)
145
148
165
Hydraulic Head
70
23
15
Pressure Head
75
125
150
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)
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
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
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.
A new well has been disinfected according to the
standard operating procedure, but a fecal
coliform sample taken after disinfection still
e.
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nf
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w
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shows colony growth. The operator should
1. Disinfect the well again without skipping any
steps.
25% 25% 25% 25%
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.
25% 25% 25% 25%
um
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...
...
fr
om
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Sl
im
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fro
m
iro
...
1. Slime from iron-oxidizing
bacteria
2. Biofilms from ammoniaoxidizing 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?
25%
25%
25%
L
ug
/
25
m
g/
L
25
L
ug
/
50
m
g/
L
50 mg/L
50 ug/L
25 mg/L
25 ug/L
50
1.
2.
3.
4.
25%
After a routine repair to an
existing well, how much chlorine
residual is required in the well to
ensure adequate disinfection?
25% 25% 25% 25%
10
00
m
g/
L
m
g/
L
40
0
m
g/
L
0
20
0
m
g/
L
100 mg/L
200 mg/L
400 mg/L
1000 mg/L
10
1.
2.
3.
4.
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
total = 740 gal, time = 5 minutes
Formula
Well yield = Flow gallons
Duration of Test, min
92%
8%
gp
m
0%
0
gp
00
37
14
8
gp
m
m
0%
gp
m
67
Solve:
Well yield = 740 gallons = 148 gpm
5 min
1. 67 gpm
2. 148 gpm
3. 3700 gpm
4. 0 gpm
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
Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft
Formula 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
79%
Solve:
65/100
14%
7%
lb
s
65
0.
2
.0
1
lb
s
lb
s
0%
lb
s
2 lbs
1 lbs
.02 lbs
0.65 lbs
2
1.
2.
3.
4.
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)
Given
Formula
0.5 Lbs/day= (X mg/L)(.000550 MG)(8.34 lb/gal)
m
g/
L
5
m
g/
L
0.
9
10
1
m
g/
L
g/
L
m
1209 mg/L
109 mg/L
1 mg/L
0. 5 mg/L
09
1.
2.
3.
4.
25% 25% 25% 25%
12
Solve:
X= 0.5/(0.00050 MG)(8.34)
X= 109 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
D=6 in=0.5 ft Well =120 ft
Formula 220 ft - 100 ft = 120 ft water in well
(0.785)(D2)(H) = ft3
(mg/L)(MG)(8.34 lb/gal)
25%
70
1.
70
5.
lb
s
gal
lb
s
07
0.
2 lbs
0.07 lbs
5.70 lbs
1.70 lbs
25%
lb
s
(0.785)(0.5 ft)(0.5 ft) (120 ft)(7.48
(50 mg/L)(.000176 MG)(8.34 lb/gal) = 0.07
lb
s
1.
2.
3.
4.
25% 25%
3
gal/ft )= 176
2
Solve:
Today’s objective: Well Disinfection, Well
Casing, Well yield and chlorine dosage of
new wells been met?
86%
Strongly Agree
Agree
Neutral
Disagree
Strongly Disagree
14%
0%
ag
re
e
0%
St
ro
ng
ly
D
Di
s
is
a
gr
ee
tra
l
eu
gr
ee
A
N
ly
Ag
re
e
0%
St
ro
ng
1.
2.
3.
4.
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