AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
1
of
7
Definition
of Sedimentation
Gravitational accumulation of solids (particles) at the bottom of a fluid (air or water)
Essentially settling of solid particles
Where is Sedimentation Used?
Removal of solids from drinking water
Most water sources, especially
surface water, have some solids that must be removed
prior to consumption
Removal of solids from waste waters
Settling of solids in a septic tank
Settling of solids in a primary lagoon
Solids removal at a treatment plant
Settling of solids from air emissio
ns
Dust deposition near a feed lot
Removal of solids from runoff water
Sedimentation basins at construction sites
Solids deposition in stock tanks
Types of Sedimentation
Discreet Settling
—
individual particles settle independently, low solids concentration
Flocculant
—
Concentration is great enough to cause individual particles to stick together and
form flocs. Flocs settle unhindered.
Hindered
—
Particle concentration is great enough to inhibit water movement between
particles during settling, water must move
through spaces between particles
Compression
—
Particles settle by compressing mass below
Influencing Factors
Particle factors
—
size, density, shape, to some extent electrical charge
Size
—
bigger particles settle faster than smaller
Density
—
denser particles s
ettle faster than less dense
Shape
—
spherical particles settle faster than large, flat particles
Charge
—
some particles (clays) carry an electric
al
charge which interacts with polarity of
water
Fluid factors
—
flow velocity, density, viscosity
Flow velocity
—
pa
rticles settle faster when there is no fluid movement, velocity causes
turbulence that keeps particles suspended (greater velocity allows suspension of larger,
denser particles)
Fluid density
—
determines amount of buoyancy (weight of displaced water), dense
r fluids
provided greater buoyancy and slower settling
Fluid viscosity
—
(fluid thickness) greater drag from thicker fluid
AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
2
of
7
Determining Sedimentation Rate
Stoke’s Law
—
applies to spherical particles settling in a very quiet fluid (usually won’t apply
to gasse
s)
μ
g
)d
ρ
(
ρ
v
w
p
p
18
2
(1)
where:
v
p
= particle settling velocity (m/s or ft/s),
ρ
p
= particle density (kg/m
3
or lb
m
/ft
3
)
ρ
w
= fluid density (kg/m
3
or lb
m
/ft
3
)
d
= particle diameter (m or ft),
g
= gravitational acceleration (9.81 m/s
2
or 32.
2 ft/s
2
)
μ
= fluid viscosity (kg/m
∙s or lb
m
/ft∙
s)
Typical values for
ρ
and
μ
Particle densities for mineral sediments (mineral fraction of soil)
ρ
p
= 2650 kg/m
3
Water density and viscosity at 20 ºC
ρ
w
= 998 kg/m
3
(62.3 lb
m
/ft
3
)
μ
= 1.01 x 10

3
kg/m∙s
6.
74 x 10

4
lb
m
/ft∙s)
(Note: values will vary with temperature
–
see attached table
)
Example
Determine the settling velocity for a 0.5mm
diameter
particle with a density of 2000 kg/m
3
in 20
ºC water.
ρ
p
= 2000 kg/m
3
ρ
w
= 998 kg/m
3
d
= 0.5mm = 0.0005 m
g
= 9.81 m/s
2
μ
= 0.00101 kg/m
∙
s
m/s
14
0
s
kg/m
00101
0
18
m/s
81
.
9
m
0005
0
kg/m
998
2000
2
2
3
.
.
.
v
p
AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
3
of
7
Application to Settling
Tanks and
Basins
The surface area of a settling tank or basin needed to trap particles of a given size and density
can be determined using Stoke’s Law.
The critical settling velocity is set eq
ual to the settling velocity of the smallest particle to be
trapped.
The overflow rate
(
OFR
) for a settling tank is
equal to the flow rate into the tank divided by
the surface area
.
This is similar to a vertical velocity.
Setting the overflow rate equal t
o the critical settling velocity will allow time for capture of
the smallest particles of interest
:
A
Q
v
OFR
c
(2)
where
OFR
= the overflow rate (m/s or ft/s)
v
c
= the critical settling velocity (m/s or ft/s)
Q
= the flow rate into the
tank
(m
3
/s or cfs)
A
= the surface area of the
tank
(m
2
or ft
2
)
Example
Particles in a
wastewater flow
ha
ve a density of
2000 kg/m
3
and range
in size from 0.2 mm to 1.5
mm. The flow rate into a settling basin is 2,000,000 m
3
/d.
What surface area
is ne
eded
for a
settling basin if it is
to capture particles 0.5 mm and larger? The temperature is 20 ºC.
We need
v
p
for 0.5mm particles which we will use as the critical velocity.
From the previous example, we know that
v
p
= 0.14 m/s for 0.5mm
diameter
parti
cle
s
having
a
density of
2000 kg/m
3
. This is the critical velocity needed which is set equal to the
OFR
:
s
3600
h
h
24
d
1
d
m
000
000
2
m/s
14
0
3
A
,
,
.
v
c
Solving for the surface area, we get
A
= 165 m
2
.
Note that we don’t get any information from
this about the depth of the basin
, only the surface area.
AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
4
of
7
Flow
Equalization
(BAEN 465)
Because t
he flow rate and strength (BOD concentration) of a wastewater stream are continually
changing
,
it is difficult to maintain
efficient treatment process operation. Flow equalization is
used to
dampen the variations in flow and
waste
strength so the wastewater can be treated at an
approximately constant flow rate. A flow equalization tank
(basin)
allows wastewater to
accumulate when flow is above average and provides additional wastewater to do
wnstream
processes when inflow is below average to maintain
a constant flow rate at
the average
value
.
Assuming the density of the wastewater remains constant, a volumetric balance can be made
around the equalization tank:
out
in
Q
Q
dt
dV
(3
)
or for a finite time interval
(Δ
t
)
t
Q
t
Q
V
out
in
(4
)
where
Q
is volumetric flow rate.
In addition to maintaining a constant flow to downstream
processes, an equalization tank will also dampen variations in t
he BOD
5
concentration in the
flow to these processes.
The best way to approach
designing an equalization tank
is a spreadsheet solution
using a table of
hourly flow rates and BOD
5
concentrations
which comprise the daily inflow cycle for the
WWTP
. Calcula
te the average flow rate then rearrange the table so the first row is the time
where the
in
flow first exceeds the average flow. The rest of the times follow in sequence to
complete the daily cycle.
C
alculate the volume accumulating in the basin as the di
fference
between inflow and outflow, which is the average flow rate, throughout the day. The
accumulated volume should increase to
a
maximum and then decrease to zero at the end of the
daily cycle.
The maximum is the volume needed for the equalization ta
nk. Additional volume
may be added to allow for equipment failures, unexpected flow variation, and solids
accumulation, e.g., the flow equalization volume requirement might be increased by 25%.
Equalization also will provide a dampening effect on fluctua
tions in the
BOD
5
loading to
downstream processes. Similar to the volume balance used above
, a mass balance on BOD
5
can
be used to determine the fluctuation in the BOD
5
concentration in the basin
,
and hence to the
WWTP, throughout the day.
The mass of
BOD
5
(
M
BOD

in
)
entering
the basin is given by the product of the inflow (
Q
in
), the
BOD
5
concentration (
S
0
) and the time interval (Δ
t
):
t
S
Q
M
0
in
in

BOD
(
5
)
The
average
BOD
5
concentration
(
S
avg
) is
determined by
adding
the
total amount of
BOD
5
contained in
the basin
at the end of
the previous period
(
V
tank

prev
S
prev
)
to
the BOD
5
that enters
duri
ng the current period
and dividing by the total volume in the basin at the end of the previous
period plus the volume entering during the current period
:
AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
5
of
7
prev

tank
in
prev
prev

tank
0
in
avg
V
t
Q
S
V
t
S
Q
S
(
6
)
The mass of
BOD
5
flowing out of the basin during the time period
is
the product of the average
flow rate, the average concentration
and the time interval:
t
S
Q
M
avg
out
out

BOD
(
7
)
Equalization can significantly reduce the fluctuation in the BOD
5
mass loading to downstream
processes.
Example
(also see Example
11

2 in the text)
The dai
ly flow cycle for a municipal WWTP
is given in the table below
using 4 h time
increments:
Flow
BOD
5
Time
(m
3
/s)
(mg/L)
0000
0.345
138
0400
0.141
89
0800
0.305
131
1200
0.498
25
1
1600
0.562
255
2000
0.550
146
Avg
0.400
The average flow rate is found to be 0.400 m
3
/s. Rearrange the table so the first row contains the
first period where the flow rate is greater than the average (1200
in this example
).
We assume
that the
equalization
tank
is empty at the beginning of this
time period.
For each row, c
alculate
the difference between the flow in and flow out (
ΔV)
then add that to the amount in the
tank
at
the beginning of that time period (Σ(V)
)
from the previous time period
. The maximum value in
the Σ(
Δ
V) column will be th
e minimum size for the equalization
tank
. The
tank
volume may be
increased to allow for equipment outages or other contingencies.
Notice that the Σ(ΔV) column
will decrease to 0
after 24 h (last row)
.
Flow
BOD
5
V
in
V
out
ΔV
Σ(ΔV)
呩浥
⡭
3
/s)
(mg/L)
(m
3
)
(m
3
)
(m
3
)
(m
3
)
1200
0.498
25
1
7171
5762
1409
1409
1600
0.562
255
8093
5762
2330
3739
2000
0.550
146
7920
5762
2158
5897
0000
0.345
138
4968
5762

794
5102
0400
0.141
89
2030
5762

3732
1370
0800
0.305
131
4392
5762

1370
0
To determine the effect
of equalization on the BOD
5
concentration in the tank, mass balances are
completed for each row. The mass of BOD
5
flowing into the basin for a time period is given by
equation
5
, and the average concentration in the tank by equation
6
. The average conce
ntration
for
the time
period is given by equation
7
.
For the first time period, the average concentration is
the same as the inflow concentration since the tank is empty at the beginning of this time period.
AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
6
of
7
Flow
BOD
5
V
in
V
out
ΔV
Σ(ΔV)
M
BOD

in
BOD
ta
n
k
Time
(m
3
/s)
(mg/L)
(m
3
)
(m
3
)
(m
3
)
(m
3
)
(kg)
(mg/L)
1200
0.498
25
1
7171
5762
1409
1409
18
00
25
1
1600
0.562
25
5
8093
5762
2330
3739
20
64
25
4
2000
0.550
146
7920
5762
2158
5897
1156
180
0000
0.345
138
4968
5762

794
5102
686
161
0400
0.141
89
2030
576
2

3732
1370
181
140
0800
0.305
131
4392
5762

1370
0
575
133
The BOD
tank
concentration is the concentration flowing to downstream processes. Notice that
the range is smaller (133

25
4
mg/L) than the range of concentrations entering the equalization
ta
nk (89

255 mg/L).
Note: In this case the maximum BOD
tank
concentration occurred during the same time period as
the maximum BOD
5
coming into the tank. This will not always be true, particularly when hourly
data are available, since the highest BOD
5
con
centrations will not necessarily occur at the same
time as the highest flow rates.
Properties of water (taken from
http://www.thermexcel.com/english/tables/eau_atm.htm
):
Temp
(°C)
De
nsity
(kg/m
3
)
Viscosity
(kg/m∙s)
Temp
(°C)
Density
(kg/m
3
)
Viscosity
(kg/m∙s)
0.00
999.82
0.001792
21.00
998.08
0.000979
1.00
999.89
0.001731
22.00
997.86
0.000955
2.00
999.94
0.001674
23.00
997.62
0.000933
3.00
999.98
0.001620
24.00
997.38
0.0009
11
4.00
1000.00
0.001569
25.00
997.13
0.000891
5.00
1000.00
0.001520
26.00
996.86
0.000871
6.00
999.99
0.001473
27.00
996.59
0.000852
7.00
999.96
0.001429
28.00
996.31
0.000833
8.00
999.91
0.001386
29.00
996.02
0.000815
9.00
999.85
0.001346
30.
00
995.71
0.000798
10.00
999.77
0.001308
31.00
995.41
0.000781
11.00
999.68
0.001271
32.00
995.09
0.000765
12.00
999.58
0.001236
33.00
994.76
0.000749
13.00
999.46
0.001202
34.00
994.43
0.000734
14.00
999.33
0.001170
35.00
994.08
0.000720
15.00
999.19
0.001139
36.00
993.73
0.000705
16.00
999.03
0.001109
37.00
993.37
0.000692
17.00
998.86
0.001081
38.00
993.00
0.000678
18.00
998.68
0.001054
39.00
992.63
0.000666
19.00
998.49
0.001028
40.00
992.25
0.000653
20.00
998.29
0.001003
AGSM 337/BAEN 465
Sedimentation, Flow Equalization
Page
7
of
7
Commentaires 0
Connectezvous pour poster un commentaire