Filter Loading and Backwash Rates
Well Yields and Chlorine Dosage in
Waterworks Operation
Math for Water Technology
MTH 082
Lecture
Chapter 4 & 8- Applied Math for Water Plant Operators
Loading Rate Calculations (pg 62-66); Well yield (163-177)
Mathematics Chapter 19-20 &23- Basic science Concepts and
Applications (pg 183-190; 197-203)
Objectives
Reading assignment:
Chapter 4 & 8- Applied Math for Water Plant Operators
Loading Rate Calculations (pg 62-66); Well yield (163-177)
Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203)
1.
2.
3.
4.
Filter overviews
Filter loading and backwash rates
Well yield
Chlorine dosage for a new well
Conventional Treatment
• Conventional Treatment – common treatment steps
used to remove turbidity from the initial source
water.
Chlorination
Ozone
UV
1. Coagulation
Raw Water
Pretreatment
2. Flocculation
3. Sedimentation
5. Clear Well
4. Filtration Disinfection
Distri
Fast
bution
Slow
Rapid or flash Mixing
Alum, polymer
Sludge
Washwater
Clearwell
Backwash
pumps
High Rate Filter
Dual or Multi Media Filter
-4 times faster then rapid sand
-granular activated carbon
-anthracite coal
-garnet sand and gravel
-backwash ready (70 hr longevity)
-excellent water quality
High Rate Filters
Anthracite Coal
Fine sand
Garnet
sand
High Rate Filter
Dualmedia Filter
Monomedia Filter
Anthracite Coal
Fine sand
Garnet
sand
Monomedia
Coarse sand
Dual-media and multimedia
filters
91%
1. Require an extremely deep bed
2. Can operate at three or four times the
rate of sand filters
3. Cannot reduce turbidity
4. Do not require backwashing
5%
eq
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tr
no
o
D
C
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t
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an
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at
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.
ex
an
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re
eq
R
5%
0%
When mixed media filters
composed of garnet, sand, and
crushed anthracite coal are
used, which of the following
describes their placement in the
filter bed?
92%
a.
..
ite
co
,a
...
nt
hr
ac
on
nd
Sa
A
n
to
ar
ne
G
0%
to
p
to
p,
co
ite
nt
hr
ac
0%
...
a.
..
8%
A
1. Anthracite coal on top, garnet in the
middle, and sand on the bottom
2. Garnet on top, anthracite coal in the
middle, and sand on the bottom
3. Sand on top, anthracite coal in the
middle, and garnet on the bottom
4. Anthracite coal on top, sand in the
middle, and garnet on the bottom
In a filter using gravel,
anthracite, and sand, the
anthracite should be?
88%
The top layer of media
Beneath the gravel
Between the sand and the gravel
Mixed with the sand
12%
th
e.
..
...
ith
w
d
ix
e
M
ee
n
et
w
B
ea
th
en
B
th
e
th
e
sa
er
la
y
to
p
e
0%
gr
...
...
0%
Th
1.
2.
3.
4.
Pilot Filters
Anthracite Coal
Fine sand
Garnet sand
The main action of a mixed
media filter is:
87%
Straining
Disinfecting
Coagulating
None of the above
ab
...
th
e
gu
l
N
on
e
of
oa
C
D
is
i
nf
ec
at
in
g
tin
g
in
in
g
9%
4%
0%
St
ra
1.
2.
3.
4.
Backwash
• Suspended particles entrapped by filter
media.
• Accumulation occurs:
– head loss within the filter to reach
excessively high levels (6 to 8 feet of
hydraulic head).
- Particles pass through the filter,
water turbidities reach unacceptable
levels
- Rule Backwash at 0.1 NTU
- SWTR Allows 0.3 NTU.
Head Loss
• Clean filter =0 psi= one foot of head loss on
a new filter
• As filter clogs more negative pressure
• Pressure builds in a linear fashion
-2.5 to -4.0 psi = 6 to 10 ft of head loss
• More clogged greater the head loss
• Remember 1 ft of water column = 0.433 psi
• 2.31 ft of water for 1 psi change
Filter Backwash
• Some plants use head loss, some use
time
• Some plants use operator knowledge
and turbidity
• Each operator has their own scheme!!
• Parents and diapers--
The most critical criterion for
determining when a mixed media
filter should be backwashed is:
Filter effluent quality
Flow rate
Head loss
Visual inspection of the filter surface
39%
Vi
su
a
H
ea
li
ns
d
pe
lo
s
s
ct
...
4%
ra
te
ow
Fl
lte
r
ef
flu
en
...
4%
Fi
1.
2.
3.
4.
52%
Filtration Rate Calculation
Unit Filter Run= (Total gal filtered gal)
Filter Area (sq ft)
Downward
Filtration Rate= (flow gpm)
Area (sq ft)
Units will be gpm
ft2!
Backwash Rate = (flow gpm)
Area (sq ft)
Units will be gpm
Upward
The total water filtered during a filter run
(between backwashes) is 2,950,000 gal. If
the filter is 15 ft by 20 ft, What is the unit
filter run volume (UFRV)?
Given
Formula
L= 15 ft, W=20 ft; Rate 2,950,000 gal
A=L X W
UFVR= (Total gallons filtered g)
Area (sq ft)
Solve:
A= 20 ft X 15 ft = 300 ft2
UFVR= (2,950,000 gal)
300 (sq ft)
UFVR = 9833 g/ft2
25%
ft2
0.
98
33
g/
33
1.
10
98
7
g/
ft2
g/
ft2
0%
g/
ft2
0%
3
9.833 g/ft2
101.7 g/ft2
9833 g/ft2
0.9833 g/ft2
9.
83
1.
2.
3.
4.
75%
The total water filtered during a filter run
(between backwashes) is 4.8 MG. If the
filter is 20 ft by 30 ft, What is the unit filter
run volume (UFRV)?
Given
Formula
L= 20 ft, W=30 ft; Rate 4.8 MG
A=L X W
UFVR= (Total gallons filtered g)
Area (sq ft)
Solve:
A= 20 ft X 30 ft = 600 ft2
UFVR= (4,800,000 gal)
600 (sq ft)
UFVR = 8000 g/ft2
8
0.
00
80
0
g/
ft2
g/
ft2
80
00
g/
ft
0%
g/
ft2
0%
2
0%
00
0
96000 g/ft2
8000 g/ft2
800 g/ft2
0.008 g/ft2
96
1.
2.
3.
4.
100%
A filter 20 ft by 25 ft receives a flow of
1940 gpm. What is the filtration rate in
gpm/ft2?
Given
Formula
L= 20 ft, W=25 ft; Rate 1940 gpm
A=L X W
Filtration Rate= (flow gpm)
Area (sq ft)
Solve:
A= 20 ft X 25 ft = 500 ft2
Filtration Rate= (1940 gpm)
500 (sq ft)
Filtration Rate = 3.9 gpm/ft2
11%
d/
ft3
gp
0.
25
gp
m
25
0.
9
3.
0%
/ft
2
m
/ft
2
6%
gp
/ft
3
gp
d
3.9 gpd/ft3
3.9 gpm/ft2
0.25 gpm/ft2
0.25 gpd/ft3
3.
9
1.
2.
3.
4.
83%
A filter 20 ft by 35 ft receives a flow of
1530 gpm. What is the filtration rate in
gpm/ft2?
Given
Formula
L= 20 ft, W=35 ft; Rate 1530 gpm
A=L X W
Filtration Rate= (flow gpm)
Area (sq ft)
Solve:
A= 20 ft X 35 ft = 700 ft2
Filtration Rate= (1530 gpm)
700 (sq ft)
Filtration Rate = 2.2 gpm/ft2
11%
d/
ft3
gp
0.
45
gp
m
0.
45
gp
2
2.
0%
/ft
2
m
/ft
3
0%
gp
m
/ft
2
2.2 gpm/ft2
2.2 gpm/ft3
0.45 gpm/ft2
0.45 gpd/ft3
2.
2
1.
2.
3.
4.
89%
A filter 25 ft by 30 ft receives a flow of 3.3
MGD. What is the filtration rate in
gpm/ft2?
L= 25 ft, W=30 ft;Rate 3.3 MG 1,000,000 gal 1 Day
D
1MG
1440 min
A=L X W
Filtration Rate= (flow gpm)
Area (sq ft)
Solve:
100%
ft2
32
1
gp
gp
d/
ft3
m
/ft
2
0%
3.
gp
m
0.
32
gp
1
0%
/ft
2
m
/ft
3
0%
3.
1.
2.
3.
4.
A= 25 ft X 30 ft = 750
Filtration Rate= (2292 gpm)
750 (sq ft)
3.1 gpm/ft3
2
Filtration
Rate
=
3.1
gpm/ft
0.32 gpm/ft2
3.1 gpm/ft2
0.32 gpd/ft3
0.
Given
Formula
A filter 25 ft by 10 ft has a backwash rate
of 3400 gpm. What is the filter backwash
rate in gpm/ft2?
Given
L= 25 ft, W=10 ft; Rate 3400 gpm
Formula A=L X W
Backwash Rate = (flow gpm)
Area (sq ft)
Solve:
100%
0%
gp
d
/ft
3
/ft
2
6
13
.6
13
4
07
0.
gp
m
gp
m
/ft
3
gp
m
.1
0%
/ft
2
0%
13
1.
2.
3.
4.
A= 25 ft X 10 ft = 250 ft2
Filter Backwash Rate= (3400 gpm)
250 (sq ft)
2
Filter Backwash
Rate
=
13.6
gpm/ft
13.1 gpm/ft3
0.074 gpm/ft2
13.6 gpm/ft2
136 gpd/ft3
A filter 20 ft by 15 ft has a backwash rate
of 4.5 MGD. What is the filter backwash
rate in gpm/ft2?
Given L= 20 ft, W=15 ft; ft;Rate 4.5 MG 1,000,000 gal 1 Day
Formula
D
1MG
1440min
A=L X W
Backwash Rate = (flow gpm)
Area (sq ft)
Solve:
94%
6%
ft3
gp
d/
01
0.
gp
m
.4
10
0.
01
gp
m
/ft
2
0%
/ft
2
0%
/ft
3
gp
m
.4
10
1.
2.
3.
4.
A= 20 ft X 15 ft = 300 ft2
Filter Backwash Rate= (3125 gpm)
300 (sq ft)
3
10.4 gpm/ft
Filter Backwash Rate = 10.4 gpm/ft2
0.01 gpm/ft2
10.4 gpm/ft2
0.01 gpd/ft3
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
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?
Given
Static WL= 140 ft, Pumped WL=167 ft
Formula
Drawdown ft = pumping water level – static water level ft
Drawdown = 167 ft- 140 ft
Solve:
Drawdown
= 27 ft
ft
0
27
-2
7
ft
0%
ft
0%
ft
0%
7
307 ft
-27 ft
27 ft
0 ft
30
1.
2.
3.
4.
100%
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
96%
4%
gp
m
0
gp
00
37
14
8
gp
m
m
0%
gp
m
0%
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
Solve:
65/100
71%
24%
6%
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.
Today’s objective: Filter Loading Rates,
Filter Backwash Rates, Well yield and
chlorine dosage of new wells been met?
Strongly Agree
Agree
Neutral
Disagree
Strongly Disagree
0%
ag
re
e
0%
St
ro
ng
ly
D
Di
s
is
a
gr
ee
tra
l
0%
eu
A
gr
ee
0%
N
ly
Ag
re
e
0%
St
ro
ng
1.
2.
3.
4.
5.