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Wastewater Engineering and Design
By Assist. Prof. Dr. Wipada Sanongraj 1
1
Chapter 2: Physical Treatment Units; Design of Sedimentation
Tanks
Wastewater Engineering and Design
2
Outline
n Physical Treatment Unit
n Types of Sedimentationn Principle of Sedimentation
n Primary Sedimentation Tank
n Settling Thickener
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Wastewater Engineering and Design
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3Figure 2-1: Physical Unit Treatment (Metcalf and Eddy,
1991)
4
Operation Applicationn Flow metering Process control, process
monitoring
and discharge reports
n Screening Removal of coarse and settleablesolids by
interception
n Flow equalization Equalization of flow and mass loading of BOD
and suspended solids
n Mixing Mixing chemicals and gases with wastewater, and
maintaining solids in suspension
n Flocculation Promotes the aggregation of small particles into
larger particles to enhance their removal by gravity
sedimentation
Physical Treatment Unit
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Operation Applicationn Sedimentation Removal of settleable
solids
and thickening of sludges
n Flotation Removal of finely divided suspended solids and
particles biological sludges
n Filtration Removal of fine resudualsuspended solids remaining
after biological or chemical treatment
n Microscreening Same as filtration
n Gas transfer Addition and removal of gases
n Volatilization Emission of volatile and semi-volatile and gas
stripping
organic compounds from WW
Physical Treatment Unit
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Types of Sedimentation
Bar Screening
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Four Types of Settling
Type I: Discrete Settling
1. Particles settle out individually and do not interact with
one another.
2. Suspension of low solids concentration.
Types of Sedimentation
8
3. Occur in:a) Grit chambers
and some primary clarifiers used in wastewater treatment
plants.
b) Some presedimentationbasins in water treatment.
Type I: Discrete Settling
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Type II: Flocculants Settling
1. As particles flocculate (due to velocity differences), they
increase in mass and velocity:
a) Mixing due to hydraulic gradients in clarifier produces
particle collisions:
10
b) Larger particles overtakes smaller ones and flocculate:
Type II: Flocculants Settling
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2. Suspension of low solids concentration3. Occurs in:
a. Most primary and all secondary clarifiers in wastewater
treatment.
b. Most sedimentation basins in water treatment.
Type II: Flocculants Settling
12
Type II: Flocculants Settling
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Type III: Zone Settling or Hindered Settling
1. Formation of dense mat of particles that settle out as a
unit:
Settling of different size particles will eventually form thick
settling blanket that settles at a constant velocity.
14
2. Suspension of intermediate concentration.3. Intraparticle
forces are sufficient enough to affect adjacent particles.4. Occurs
in:
a. Thickeners (sludge disposal)b. Bottom of clarifiers and
sometimes
sedimentation basins.
Type III: Zone Settling or Hindered Settling
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Type IV: Compression Settling
1. Settling occurs by compression where water is forced out of
the interstitial voids between particles.
In order to settle, water must pass through particles arranged
like porous media. Furthermore, there is a high head loss due to a
large surface to volume ratio. Therefore, settling occurs
slowly.
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2. Suspension by very high concentration.3. All particles are
influenced by the presence of other particles.4. Occurs in drying
beds and some filtration process.
Type IV: Compression Settling
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Grit Materials
Sand, gravel, egg shells, coffee grounds, fruit rinds, seeds,
bones (whos)
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Principles of Sedimentation
Vp = Volume of particles = 1/6 Dp3 (cm3)Ap = Projected surface
area of particle = /4Dp2 (cm2) Fp = Drag force = friction factor *
inertial force (g-cm/s2)Fb = Bouyant force (g-cm/s2)Fg = Force due
to gravity (g-cm/s2)Vs = Velocity of particle (cm/s)mp = Mass of
particle (g)g = Acceleration due to gravity (cm/s2)rl = Density of
water (g/cm3)rp = Density of particle (g/cm3)
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Momentum Balance
( )dbg
sp FFFFdt
Vmd--==
( )2
2s
pldplppsp VACgVgV
dt
Vmdrrr --=
Determination of the Drag Coefficient in water
Cd = f(NR) & Particle Shape
Cd can be determined if the pressure and shear stress
characteristics surrounding an object are known. However, Cd can
also be determined if the total drag is measured by a force
dynamometer.
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Principles of Sedimentation
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Larminar region: viscous forces control the drag forceTurbulent
region: inertial forces of displaced fluid are controlling the drag
force
For laminar flow: (NR < 1)
l
splR
VDN
mr
=
Rd N
C24
=
splspl
spl
lspldd VD
VD
VDV
ACF pmpr
rmr 3
2*424
2
222
=
==
Principles of Sedimentation
22
Remembering the general equation:
( )splplpp
sp VDgVgVdt
Vdm pmrr 3--=
( )p
spl
p
pl
p
pps
m
VD
m
gV
m
gV
dt
Vd pmrr 3--=
Substituting for mp and Vp:
( )
pp
spl
pp
pl
pp
pps
D
VD
V
gV
V
gV
dt
Vd
rppm
rr
rr
6
33
--=
( )pp
sl
p
lps
D
Vg
dt
Vd
rm
rrr
2
18-
-=
Principles of Sedimentation
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Let l
ppD
mr
t18
2
=
( )tr
r sp
ls Vgdt
Vd-
-= 1
Using the integration factor method and the initial conditions:
Vs = 0 @ t = 0
( )t /lsp
V g 1 1 e- t r
= t - - r Where the relative measure of the fractional approach
to SS is:
( )t /1 e- t-
Principles of Sedimentation
24
l
ppDtimerelaxationmr
t18
2
==
When t then VsVtl
tp
V g 1 r
= t - r
( )t /s tV V 1 e- t= -ornow (1-e-3) = 0.95 (it takes 3t to
attain 0.95Vt)
Principles of Sedimentation
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25
Example 2-1: Typical grit particle (sand)
Dp = 100 mm = 0.01 cmrp = 2.65 g/cm3
Since the typical retention time for a particle may be on the
order of hours or more, we are not concerned about non-steady
state.
26
Therefore, we can assume steady-state:
( ) p ls l s2
p p p
d V 18 V0 g
dt D
r -r m= = - r r
Solve for Vs: (Stokes law for settling)2
p l ps R
l
g( )DV for N 1
18
r -r=
m
Example 2-1: Typical grit particle (sand)
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Example 2-2: Bio floc particle
ml = 0.01185 g/(cm-s) rp = 1.05 g/cm3 rl = 1.0 g/cm3Dp = 100 mm
T = 25 oC
Check NR
Since NR < 1, laminar flow exists and Stokes Law is Valid
28
What about a large particle size (sand)?
rp = 2.65 g/cm3 Dp = 200 mm T = 25 oC
Since the Reynolds number is greater than 1.0, Stokes law is not
valid because the drag coefficient of the particle in water does
not vary as 24/NR, but as:
Example 2-2: Bio floc particle
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Rearrange the original force balance:
( ) 2p s sp p l p d l p
d m V VV g V g C A 0
dt 2= r -r - r =
Solving for Vs yield the following equation for Newtons law for
settling particles:
Example 2-2: Bio floc particle
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Using NR = 5.11 from above:
d
24 3C 0.34 6.36
5.11 5.11= + + =
recalculate NR NR = 4.38
Example 2-2: Bio floc particle
Since we cannot solve explicitly for Vs as in Stokess Law
expression, we must now use a trial and error solution as
demonstrated below:
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The procedure for calculating the settling velocity of a
particle when NR > 1 is:
1. Use Stokes Law to obtain the initial guess for VS.
2. Use this Vs to obtain NR, if NR > 1, then go to next
step.
3. Use the following iteration process:
a) Use NR to get Cd
dR R
24 3C 0.34
N N= + +
32
c) Use Vs to find new NRd) Repeat until convergence is met.
b) Use Cd to get new Vs
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Brownian Motion
In water and WW, colloidal particles are usually present and are
hard to remove by settling. A colloidal particle may be thought as
a giant molecule and its Brownian motion is really a diffusion
process, where diffusion results from the random thermal motion
(due to thermal gradients within the fluid).
34
The above particle motion was first observed for water molecules
by Dr. Robert Brown. In 1905, Einstein used Stokess relationship
and developed the following equation to describe the motion.
Brownian Motion
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Bp
1 2kTV
X 3 D
= D pm
Where: k = Boltzman constant = 1.38*10-16 (g-cm2)/(s2-K)
T = temperature (K) DX = net x component distance of travel
(cm)m= kinematic fluid viscosity (g/cm-s)Dp = mean particle
diameter (cm)VB = particle velocity due to Brownian Motion
(cm/s)
When VB > Vs (stokes), particles will not settle out of
solution because the motion of the particle is governed by
collision withthe water molecules.
Brownian Motion
36
The following characteristics are typical for water and
wastewater colloids:
Table 2-1: Characteristics of water and wastewater colloids
(Metcalf & Eddy, 1991)
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Example 2-3:
Calculate the smallest settleable sand particle in water at 25
oC.
Assume DX = 1.0 cm
Solution
38
Setting VB = Vs
Coagulation and flocculation can be used to bring particles
together to form larger ones that will settle out of solution by
gravitational forces.
Example 2-3:
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Characteristics of Colloidal Particles in Water
1) Size-Very Small
1-1000 nm10-10,000 Angstroms10-4 10-7 cm
2) Surface Area Very Large
Diameter of Spheres Surface Area1 cm 0.0487 in2
10-4 cm 33.8 ft2
10-8 cm 3.8 yd2
10-6 cm 0.7 acres10-7 cm 7.0 acres
3) Charge Colloidal particles are usually NEGATIVELY charge
causing repulsion of similar charge and colloidal stability.
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Design of Type I Sedimentation Basins
Assumptions:1. Plug flow exists in settling zone where fluid
moves across the settling zone with a constant velocity.
2. Designed to remove discrete particle (settling zone only
considered)
3. Particles uniformly distributed at inlet.
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Key Process Variables
1. Hydraulic retention time = t = V/Q = Vol. tank/Flowrate2.
Overflow rate = Vo =Q/Atop
Vs = Flowrate/Top cross sectional area = h0/t
Vf = horizontal fluid velocity ho = height of settling zoneL =
Length of settling zone
If VsVo, all particle of that diameter are removed.If Vs
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Say Dp = 200 mm, Vs = 0.0235 m/s
Will all the particles be removed?
Example 2-4:
44
Say Dp = 100 mm, Vs = 0.0076 m/s
l
l
Example 2-4:
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Removal efficiency is independent of depth (ho) and retention
time (t)
Vs = f(g,ml,rl, rp,Dp) Vo = f(Q, Atop)How can we use this
information to determine the R.E. of a sedimentation tank with many
sizes of particles?
(Metcalf & Eddy, 1991)
46
Break the problem into two parts.
1. For particles which have Vs Vo all particles are removed or
(1-Xo) is removed.
2. For Vs Vo a fraction of the particles with such velocities
are removed.
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Break down the curve for all particles with settling velocities
Vs
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How do we obtain the necessary information to make this
calculation?
Two ways:1. Use Stokes equation or Newtons equation
depending on NR2. Use a column settling test (most common)
1. Put a suspension in column.
2. Mix well and sample to determine Co.
3. Stop mixing and begin timing.
4. Take samples from same depth.
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Example 2-5:For the following data and an overflow rate of 2.0
m/hr, what is the R.E.?
Solution1. Plot x vs. Vs2. Determine Xo from overflow rate (Vo)
3. Determine R.E. from
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ox
o so o
1R .E. (1 X ) V dx
V= - +
Example 2-5:
52
Now suppose youre given R.E. and you need to calculate Vo.
Solution: 1. Plot x vs. Vs2. Assume an Xo3. Go to curve and get
Vo4. Calculate:
Example 2-5:
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5. Calculate R.E.
6. Plot R.E. vs. Vo7. Use graph to get overflow for the R.E. of
interest.
Example 2-5:
54
Example 2-6:
For the following data, calculate the overflow rate for a R.E.
of 75%.
ox
o so o
1R.E. (1 X ) V dx
V= - +
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Solution1. Assume Xo = 0.207
2. Assume Xo = 0. 417
Example 2-6:
56
3. Assume Xo = 0. 65
Example 2-6:
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58Figure 2-1: Two approaches to determining minimum particle
size removed in a nonturbulent chamber (Mihelcic, 1999)
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Type I Sedimentation for a Circular Clarifier
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But Vf = Q/SA = Q/2prH = area of a cylinder with height H
All particles with Vs > Vo will be 100% removed. The analysis
is the same for both circular and rectangular tanks employing type
I sedimentation.
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61Figure 2-2: Typical rectangular primary sedimentation tank
(Metcalf & Eddy, 1991)
62
Horizontal Flow Grit Chambers
Flow control weirs typically require free discharge and hence a
relatively high head loss (typically 30 40% of flow depth)
Proportional weirs may cause higher velocities at the bottom
leading to bottom scour
Where effective flow control is not achieved, channels will
remove significant quantities of organic material requiring grit
washing and classifying
With effective flow control, removal of grit not requiring
further classification is possible
Excessive wear on submerged chain and flight equipment and
bearings
No unusual construction is required
Difficulty in maintaining a 0.3 m/s velocity over a wide range
of flows
Flexibility to alter performance is possible by adjusting outlet
control device
DisadvantagesAdvantages
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Aerated Grit Chambers
Flexibility to remove grit can adapt to varying field
conditions
Significant quantities of potentially harmful volatile organic
and odors may be released from wastewaters containing these
constituents
Pre-aeration may alleviate septic conditions in the incoming
wastewater to improve performance of downstream treatment units
Aerated grit chambers ca also be used for chemical addition,
mixing, pre-aeration, and flocculation ahead of primary
treatment
Some confusion exits about design criteria necessary to achieve
a good spiral roll pattern removal system
By controlling the rate of aeration, a grit of relatively low
putrescible organic content can be removed
Additional labor is required for maintenance and control of the
aeration system
Head loss through the grit chamber is minimal
Power consumption is higher than other grit removal
processes
The same efficiency of grit removal is possible over wide flow
range
DisadvantagesAdvantages
64
Optimum velocity is 3 m/s0.15-0.4Horizontal velocity, m/s
Function of velocity and channel length
15-90Detention time at peak flow, s
Based on theoretical length20-50Allowance for inlet and outlet
turbulence, %
Function of channel depth and grit velocity
3-25Length, m
Depend on channel area and flow rate
0.6-1.5Water depth, m
Dimension
CommentRangeItem
Table 2-2: Horizontal Flow Grit Chambers Typical Design Criteria
(Metcalf& Eddy, 1999)
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Table 2-3: Aerated Grit Chambers Typical Design Criteria
(Metcalf & Eddy, 1999)
Provide valves and flow meters to allow proper adjustment
0.6 0.75Transverse roll velocity, m/s
0.45 typical0.27 0.74
Medium to coarse bubble
Air supply m3/min-m
Type of diffuser
3 typical2 5 Minimum detention time (at peak flow), min
Varies widely
4: 1 typical
1.5:1 typical
0.2-5
3:1 5:1
1.1 5.1
Dimensions
Depth, m
L: W ratio
W: D ratio
CommentRangeItem
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Type II: Primary Sedimentation Tanks
Primary sedimentation tanks are designed to:1. Reduce solids
loading to minimize operational
problems in downstream treatment processes.2. Lower the oxygen
demand.3. Decrease the rate of energy consumption for
oxidation of particulate matter.
These effects enhance soluble substrate removal during aeration
and reduce the volume of waste activated sludge that is generated.
Also removes floating material, thereby minimizing operational
problems in downstream treatment processes (scum build-up in
secondary treatment processes. If efficiently designed, removal
rates of 50 to 70% of suspended solids and 25 to 40% of BOD5 can be
realized.
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Flocculent Settling
Depends Upon:1. Overflow Rate2. Depth of Basin3. Velocity
Gradients in the Basin4. Concentration of Particles5. Range of
Particle Sizes6. Settling Column Analysis of
Flocculating Particles.
68
Example 2-7:
A column analysis of flocculating suspension is performed in the
settling column shown below. The initial solids concentration is
250 mg/L. The data is also shown below. What will be the overall
removal efficiency of a settling basin that is 3 meters in depth
with a retention time of 1 hour and 45 minutes
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Solution:1) Determine the removal rate at each depth and
time.
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2) Plot iso-concentration lines as shown is the accompanying
figure.3) Construct vertical line at to = 105 min.4) Calculate the
percentage removal using the following equation:
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71
72
This procedure is very cumbersome and time consuming. Another
easier method is to calculate the removal efficiency from the
average concentration in the column. For a given retention time, t,
the average concentration in the column can be calculated using the
following equation:
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73
74
Now add 1 to both side of the last equation in the previous
slide.
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Type III Settling - Thickening
Definition: a process for removing water from sludge which
produces a more concentrated slurry (i.e. end product is a
liquid).
This process is not dewatering which produces end product with
the properties of solid.
Importance of Thickening:1. Produce a greater volume of product
water.2. Produce a smaller volume of sludge.3. Reduces the size of
a sludge treatment facility,
(e.g. digester).
78
Purposes of Sedimentation Basins:
1. Clarification
2. Thickening
With respect to design of sedimentation basins (or clarifiers)
it is essential to consider both of these functions.
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79(Metcalf & Eddy, 1991)
80(Metcalf & Eddy, 1991)
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Batch Settling Test
(Metcalf & Eddy, 1991)
82
Design Considerations for Thickening of Sludge
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In which,Q = influent flow rate to the clarifier/thickener,
(L3/t)Co = influent suspended solids concentration, (M/L3)Ce =
effluent suspended solids concentration, (M/L3)Ac = area available
for clarification, (L2)AT = area available for thickening, (L2)Qu =
flow rate leaving bottom of the
clarifier/thickener, (L3/t)Cu = suspended solids concentration
leaving the
bottom of the clarifier/thickener, (M/L3)Vi = settling velocity
of the suspended solids, (L/t)Ci = suspended solids concentration
of the blanket,
(M/L3)Ub = velocity of the solids in the underflow due to
pumping (L/t)
84
Assume: Ce
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In which,GT = total solids flux in the clarifier (M/L2-t)Gg =
settling flux in the clarifier due to gravity (M/L2-t)Gb = bulk
flux due to under flow pumping (M/L2-t)
2) Sizing a clarifier for thickening based on solids flux
method.
( ).T
masssolidsG Totalflux
area time=
In which, area is equal to the cross sectional area of the
sludge blanket
GT = Gg + Gb
86
At the bottom of the clarifier, Ci = Cu so the expression for Gb
can be written as:
u ub b u
T
Q CG U C
A= =
Since the product of QuCu is usually unknown, it can be replaced
by QCo from the above mass balance, QCo = QuCu.
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87
In addition, at the bottom of the clarifier, Gg = 0, therefore,
the area for thickening can be rewritten as:
u u o oT
L L L L b
Q C QC QCA
G G C (V U )= = =
+
88(Metcalf & Eddy, 1991)
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89
(Metcalf & Eddy, 1991)
90
(Metcalf & Eddy, 1991)
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91
(Metcalf & Eddy, 1991)
92
Alternative definition sketch for the analysis of settling data
using the solids-flux method of analysis (Metcalf & Eddy,
1991).
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93
Example 2-8:
Given the treatment scheme, estimate the maximum concentration
of the aerator mixed-liquor biological suspended solids
concentration that can be maintained if the sedimentation tank
application rate is fixed at 600 gal/ft2-d and the sludge recycle
rate, Qr, is 40% of Q. The following settling data was obtained
from operation of a pilot plant.
94
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95
Solution1. Develop the gravity solids-flux curve from the given
data.
96
2. Determine the underflow bulk velocity and plot the curve on
the same graph as the gravity solids flux curve.
b 2
3
0.4Q gal 1 ftU 600 0.95
gal hQ 0.4Q ft d h7.48 24ft d
= = + -
Ub is the slope of the curve for the underflow solids flux. The
underflow flux curve can be plotted using the following
equation:
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97
b i bG kC U=
In which,
98
3. Develop the total flux curve for the system by summing the
gravity flux curve with the underflow flux curve, and determine the
value of the limiting flux and the maximum underflow
concentration.
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99
21,800 /R uC C mg L= =
100
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101
Example 2-9:
102
Given the settling data in the following table derived from an
activated sludge pilot plant, determine the limiting solids flux
values when the concentration of the recycled solids concentration
is 10,500 and 15,000 mg/L, respectively. Determine the recycle rate
to the sedimentation tank required for thickening in conjunction
with the aeration tank, if the MLSS in the aeration tank is to be
maintained at 5,000 mg/L and the underflow concentration from the
sedimentation tank is 12,000 mg/L. Neglect the effect of biological
growth in the sedimentation tank and assume the flow is equal to
1.0 MGD.
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103
104
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105
106
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107
10,500
108For an underflow concentration of 12,000 mg/L the limiting
solidflux, SFL, = 1.8 lb/ft2-h
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109
110
22
(1 0.71)(1.0 )(5,000 )8.34 1,650
(1.8 / )(24 / )
lbmgMGD
MGLA ftmglb ft hr hr dL
+= =
-
The corresponding surface hydraulic loading rate is 606
gal/ft2-d.
-
Wastewater Engineering and Design
By Assist. Prof. Dr. Wipada Sanongraj 56
111
Table 2-4: Typical Design Information for Secondary
Clarifiers(Metcalf & Eddy, 1999)