Notes 2b A4

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AGSM 337/BAEN 465


Sedimentation, Flow Equalization



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1

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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



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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



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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



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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


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



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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



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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



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