CE 326 Principles of Environmental Engineering
Sedimentation Basins
Dr. S.K. Ong
Purpose – to physically separate solid materials from water by gravity
Typical sedimentation basins or tanks, clarifiers, settling tanks can be categorized according to water flow:
- _____________________ flow
- _____________________ type
Sedimentation tanks may be circular, square or rectangular. Depths of sedimentation tanks vary
________________. For circular tanks the diameters are from ____________ feet while rectangular tanks are
___________ feet wide and up to ______________ feet long.
Upflow
Horizontal
Ideal Settling Tank
Has four zones
- _____________
- _____________
- _____________
- _____________
Assumptions
- steady flow
- when particles hot the sludge zone - remain there
- flow through period = detention time (no dead zones)
- settling in discrete particles
- move forward with the same velocity as the water
9
To simply analysis, settling may be classified as:
__________ Settling
free settling, settling of particles as separate units and there is no apparent interaction
between particles. Examples: presedimentation and settling of sand
__________ Settling
The particles flocculate during settling and increase in size and settle at a faster velocity.
Examples: settling of coagulated waters, primary settling of municipal
wastewater.
__________ Settling
zone settling or hindered settling. Settling of an intermediate concentration of particles
where particles are so close together that interparticle forces hinder the settling of
neighboring particles. Particles settle at a constant velocity. Therefore there will be a
distinct solid-liquid interface between the settling particle mass and the clarified liquid.
Examples: settling in the intermediate depths of water treatment clarifiers and
wastewater treatment plant secondary clarifiers.
___________ Settling
compression settling. Settling occur by compression of the compacting mass and
interparticle liquid is squeezed out. Settling of particles of high concentrations.
Examples: compression settling in lower depths of thickening clarifier for activated
sludge plants
Estimation of settling velocity of a particle (Type 1 settling only)
A particle settling in water is subjected to the following forces:
FG = gravitational force
FB = Buoyancy force
FD = Drag force
When upward and downward forces are equal, particle will settle at a constant velocity called terminal velocity.
Equating forces:
_______________________________
FG = _________________
FB = _______________
s = density of particle (kg/m3)
g = gravitational acceleration (m/s2)
VP = volume of particle (m3)
 = density of fluid/water (kg/m3)
CD = Drag coefficient
AP = cross sectional area of particle (m2)
Vs = terminal velocity (m/s)
s g Vp -  g Vp = CD Ap  (Vs2/2)
Assume the particle is spherical with a diameter d:
VP = (/6) d3
AP = (/4) d2
9
FD = ___________________

d3
VP
2
6

 d
AP  d 2 3
4
Then
CD is dependent on the flow regime and is given by:
Laminar flow
Re < 0.5
CD = 24/Re
Transition flow
0.5 < Re < 10,000
CD = 24/Re + 3/Re½ + 0.34
Turbulent flow
Re > 10,000
CD = 0.4
Re = Reynolds number =  d Vs/
where
 = kinetic viscosity (m2/s)
 = shape factor, to describe the nonspherical shape of the particle
For particles falling under laminar conditions,
CD = 24/Re
= 24 / ( d Vs/) = 24  / d Vs
Therefore
Stoke’s Equation
where  = dynamic viscosity (Pa s-1)
Typical Upflow Clarifier:
Vs - rate at which the particle is settling
Vo – flow of water upwards
If Vs > Vo – ______________________
If Vs < Vo – ______________________
If Vs = Vo – _____________________
Because of flow of water is not uniform and there is a distribution of different particles and densities, , Vo is taken as
< 80% of Vs in most settling tanks. Note that Vs is usually not known.
9
Vo (the average flow rate of the water) is given by:
flow rate /surface area of the clarifier = ______________
m3/d/m2 = m/d (velocity unit)
where As is the surface area of the clarifier
Vo is termed as surface loading area (amount of flow per unit area) or overflow rate
Note that the overflow rate for a horizontal clarifier is computed using the above equation.
Design parameters - Summary
- overflow rate or surface loading rate or clarification rate (m3/s/m2)
- weir overflow rate (m3/d/m)
Others
- particle settling velocity Vs (m/s)
- detention rate (hr)
Notes
 the overflow rate is the average fluid velocity at about the mid section of the sedimentation tank. At the collection
weir at the edge of the clarifier, the fluid velocity may be higher, resulting in a carry over of the particles. Therefore
sufficient length must be provided to ensure that water will not rush over the weir. This lead to the second design
parameter, weir overflow rate given by:
= flow rate/length of weir = _______________
 there is a distribution of particle sizes therefore will have a range or distribution of V s. Sedimentation tanks are
usually not designed based on the settling velocities of the particle.
 Detention Time - approximately 2 - 8 hours (4 - 6 hours common)- some lime softening system has a retention
time of 2 hours.
Application
Alum
Iron Salts
Lime
Overflow Rate (gal/d/ft2)
Weir Overflow Rate (gal/day/ft)
9
Sand Filters
Purpose - used primarily to remove suspended particulate materials from water, example, silt and clay,
microorganisms, colloidal and precipitated humic substances, alum or iron, calcium carbonate and magnesium
hydroxide from lime softening, iron and manganese precipitates.
Types of filters
_________________ - solids are removed within the granular materials, example rapid granular-bed filters
_________________ - solids are removed on the entering face of the granular material or filtering material,
example, slow sand filters, membrane filters
Design Parameters for rapid granular-bed filters
1. Loading rate or filtration rate, filtration velocity, V a = ______________
Where Q
As
= flow rate (m3/d)
= surface area of filter (m2)
Rapid Sand Filter Loading rates
 Conventional pretreatment using alum and iron salts/sand medium
 High rate with dual or triple media (anthracite coal layer)
______ gpm/ft2 (_________ m/h)
_______gpm/ft2 (_________ m/h)
2.
Practical maximum area of an individual gravity bed filter ____________ m2.
3.
Minimum number of individual filter beds - two. When one is backwashing or under maintenance - the
other bed must be capable of meeting water demands.
4.
Total depth of and size of media
Single media
Dual media
5.
graded sand
Anthracite and sand
0.5 to 0.75 m
Sand = 0.3 m, anthracite = 0.45 m
Headloss calculations through the sand filter (equation , page 426)
where
hL = head loss through the filter, m
Va = filtration rate m/s
D = depth of filter sand, m
fi = mass fraction of sand particles of diameter, d i
di = diameter of sand grains of fraction i, m
e = porosity
 = shape factor
CD = drag coefficient
 CD
f
fi
f
f
 C D 1  C D 2  C D 3  .......... ..
di
d1
d2
d3
Initial head loss through media < 0.6 m
Terminal head loss through media between 1.8 to 2.4 m
9
Note that the sand used usually has a range of sizes. To compute the head loss the sieve analysis (size
distribution) is needed. If the sieve analysis is provided by the supplier use the sieve analysis (size
distribution) by the supplier. However, some suppliers provide:
Effective size, P10 – 10th percentile – size of sieve opening where 10% of the total mass would go
through
Uniformity Coefficient , U = P60/P10
The size distribution curve of the sand can be described by the following equation:
Px = Xg (Sg)z
where
z = normal variate of the log normal curve (see Table below)
x = percent of the fraction passing through the sieve
Xg = geometric mean
Sg = geometric standard deviation
Fraction
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Percent
10
20
30
40
50
60
70
80
90
Z
-1.282
-0.842
-0.524
-0.253
0.000
0.253
0.524
0.842
1.282
Using P10 and U, Xg and Sg can be estimated using the following equations. With Xg and Sg the geometric
distribution of the sand can be estimated.
P10 = Xg (Sg)-1.282
U = P60/P10 = (Sg)1.535
Example: a sand has the following specifications, P10 = 0.40 mm, U = 1.40
Sg = (1.4)1/1.535 = 1.245
Xg = 0.40/(1.245)-1.282 = 0.530 mm
For typical filter head loss calculations, the sand is divided into five different fractions (20% each).
Therefore
P30
= (0.530)(1.245)-0.524
= 0.472 mm
9
Problem 3 - 81 - Find head loss, given Depth D = 0.75 m, Ss = 2.80, = 0.91, e = 0.50, T = 5o C, Vs = 230 m3/d.m2
Step 1.
Compute the filtration rate, in this problem it is given as Vs = 230 m3/d.m2 = _________________ m/s
Step 2.
Determine the geometric mean of sand particle for various fractions, for example, fraction retained by US sieve 8 - 12, d = (2.38)(1.68) = ___ mm
Step 3.
Determine Reynolds number Re 
Step 4.
Compute CD, transition range use C D 
Step 5.
Compute (CDfi/di) and sum all values
Step. 6.
Compute head loss using hL 
US Standard
Sieve No
8 – 12
12 – 16
16 – 20
20 – 30
30 – 40
40 – 50
50 – 70
% retained
0
0.40
13.10
54.40
30.20
1.785
0.015
dvs (0.91)( 0.002 )( 2.66 x10 3 )

 3.70

1.307 x10 6
24
3
24
3

 0.34 

 0.34  8.38
Re
3.70
Re
3.70
1.067 s2 D
g 4
Sieve opening
(upper) mm
2.38
1.68
1.19
0.84
0.59
0.42
0.297

C D f 1.067 (2.66 x10 3 ) 2 (0.75)

38628 .37  0.39 m
d
(0.91)(9.8)( 0.50 ) 4
Sieve opening
(lower) mm
1.68
1.19
0.84
0.59
0.42
0.297
0.210
Geometric mean
(mm)
2.00
1.41
1.00
0.70
0.50
0.353
0.250
9
R
CD
3.704
2.611
1.852
1.296
0.926
0.654
0.463
8.378
11.39
15.5
21.49
29.37
40.75
56.6
CDf
d
0
32.31
2030.5
16731.5
17739.5
2060.6
33.96
38628.37
6.
Backwash Rate
- backwash water needed to clean the filters, water may be provided from the clear well or from a separate
backwash water tank
- different approaches in estimating the backwash rate
- typically _____________________________ m/d (___________________ gpm/ft2)
- about 5 times more than the filter loading rates
- expand bed by _______________________%
- wash time is about ___________________ minutes, total time backwash filter (including preparation,
draining, etc.) may take about _______________ minutes
- for typical systems, amount of backwash water used should be between _____________% of filtered
water produced
- surface wash system – to ensure thorough cleaning and to prevent the formation of mud balls – provided
for by directing jets of water downward on the surface of the filter media or by using rotating arms that
sweep the surface of the filter.
- air wash system – air introduced into the underdrain at a rate of ________________ m/min based on filter
area. The rolling action of the air scours the deposits on the sand – releasing them.
Backwash Rate
Book’s approach - a back wash rate is selected which is equal to the settling velocity of the smallest particle
that must be retained:
where
De = depth of the expanded bed, m
e = porosity of the bed
ee = porosity of expanded bed
f i= mass fraction of sand with expanded porosity
vb = backwash water velocity
vs = settling velocity
For a typical media of different sizes, divide the media into five different sizes and compute D ei for each
size and sum up the bed expansion.
7.
Placement of Backwash Trough
- bottom of backwash trough should locate about _____________ m above the expanded bed, spacing of
troughs approximately _____________ m.
Rule of thumb
- allow approximately _______________ m above the top of the sand for washwater troughs
8. Underdrain System (see Figure 3-40):
Manifold and lateral underdrain system
- first and oldest, not common now
- high head loss
- graded gravel needed
Self supporting underdrain system
- grouted to the filter floor
- vitrified clay block
False floor underdrain system
- concrete floor or steel plate (usually in small filters or pressure filters)
- located 1 to 2 feet above the bottom of the filter
- nozzles are used to collect the water and the fine openings retain filter medium
- no need for grout
9
Problem 3-82 - Find backwash rate and height of Backwash trough needed.
Step 1.
Determine the maximum backwash rate needed
- must retain the finest sand of diameter 0.210 mm (0.025 cm or 0.00025 m) and
density Ss = 2.80, from Figure 3 - 35,
minimum settling velocity vs= 1.25 cm/sec (0.0125 m/s) or 1080 m/d
Fraction
size (m)
Estimated Vs
(cm/s) (Step 2)
R
(Step 3)
CD
(Step 4)
vs
(Step 5)
ee
(Step 6)
f/(1-e)
0
0.004
0.1315
0.545
0.302
0.0178
0.00015
0.002
0.0014
0.001
0.0007
0.0005
0.00035
0.00025
20
18
15
10
7
5
3
278.5
175.4
104.4
49.7
24.4
12.2
5.22
0.61
0.72
0.86
1.26
1.93
3.17
6.25
0.185
0.14
0.11
0.076
0.052
0.0344
0.020
0.55
0.588
0.620
0.672
0.73
0.80
0.90
(step 7)
0
0.0089
0.246
1.66
1.118
0.089
1.5 x 10-5
3.22
Step 2.
Step 3
Step 4
Estimate settling velocities of each fraction of sand using Figure 3 - 35
dvs (0.91)(0.002 )(0.20)
Estimated Reynolds number using Re 

 278 .5

1.307 x10 6
24
3
24
3

 0.34 

 0.34  0.61
Compute CD , transition range use C D 
Re
278 .5
Re
278 .5
if in laminar or turbulence range use different equations
1/ 2
Step 5
Compute new vs using  4g(s  )d 


 3C D  
0.22
Step 6
Step 7
Step 8
Step 9
v 
 0.0125 
Compute ee   b 


v
 0.185 
 s
f
Compute
, sum up values
1  ee
1/ 2
 4(9.8)( 2,800  1,000 )0.002 


3(0.61)1,000


= 0.185 m/s
0.22
 0.55
f
 (1  0.5)( 0.75 )(3.22 )  1.21m
1 e
Use a safety margin of 0.15 m, therefore the total height from the bottom of the trough to the bottom of the sand = 1.21 + 0.15 m = 1.36 m.
Compute De from (1  e) D 
9
Disinfectants
Chlorine
CE 326 Principles of Environmental Engineering
Disinfectants
Reactions
CT values
Reacts with water to form hypochlorous acid
Cl2 + H2O <==> HOCl + H+ + ClHOCl <==> OCl- + H+
(HOCl - 100 times more effective than OCl-)
In the presence of ammonia, forms
chloramines
NH3 + HOCl => NH2Cl + H2O
monochloramine
NH2Cl + HOCl =>NHCl2 + H2O
dichloramine
NHCl2 + HOCl => NCl3 + H2O
trichloramine
2NH3 + 3HOCl =>N2(g) + 3HCl +
3H2O
Advantages/Disadvantages
free available
chlorine between
70 - 200 for 99.9
% inactivation of Giardia
depending on pH,
temp., concentration
- very strong oxidant
- fairly inexpensive, easy to
store and
transport
- chlorine concentration can be
determined easily
- persists in water
- easy to store
CT value for
chlor-amines is
approx. 1,850
- forms trihalomethanes (THMs)
- poisonous gas - must be
handled
carefully
- imparts odor to water at high
concentrations
2.5 for 99.9 %
inactivat-ion of
Giardia
at 10˚ C
- very strong oxidant
- leaves no residual, forms O2
- does not form THMs
Free available residual chlorine
combined available residual chlorine
Ozone (O3)
Ozone reacts with water to form OH• hydroxyl radicals, very strong oxidizing agent
There are many reactions one of them is as
follows:
- expensive, generated on site by
passing
air or O2 through a corona
discharge
generated at high voltage
- not persistent in water
O3 + OH- => O3- + OH•
9
Chlorine
dioxide
(ClO2)
- generated on site using chlorine and sodium
chlorite
Cl2 + 2NaClO2 => 2ClO2 + 2 NaCl
40 for 99.9 %
inactivat-ion of
Giardia
at 10˚ C
- does not form THMs
- more effective than chlorine
and over a
broad pH range
- provides a measurable residual
- forms chlorite and chlorate in
water,
possible health risk
- moderately expensive
compared to
chlorine
- leaves tastes and odors
Ultra Violet
(UV)
wavelength in the range of 0.2 to 0.39 µm,
lamps protected by a quartz sleeve
- no chemical addition
- performs well for bacteria and
viruses
- penetrates only 50 to 80 mm,
needs
multiple tubes
- expensive
- no residual
- water must be free from
turbidity,
lamps must be free from slimes
and
precipitation
9
CE 326 Water Treatment Plant Flow Diagram
Raw
Water
Rapid
Mix
Important Design Criteria
(i) ___________________
(ii) __________________
Slow Mix/
Flocculation Basin
Sedimentation Tank/Clarifier
Important Design Criteria
(i) ___________________
(ii) __________________
Important Design Criteria
(i) ___________________
(ii) __________________
Sand Filter
Distribution
Storage and
disinfectant
contact tank
Important Design Criteria
(i) ___________________
(ii) __________________
(iii)__________________
Important Design Criteria
(i) ___________________
(ii) __________________
Supernatant back
to rapid mix tank
Sludge Settling
Tank
Sludge
9