Primary Sedimentation tank
Primary Sedimentation tank
Purpose:
1- Removal of 40 - 60 % of suspended solids 2- Removal of 25 -
35 % of B.O.D. 3- Sediment the organic and inorganic matters to
improve the properties of the sewage and prepare it for the
biological treatment.
Types of primary sedimentation tanks:
1- Rectangular tank.
2- Circular tank.
Factors affecting sedimentation efficiency:
1- Viscosity 2- Concentration of suspended solids 3- Retention
period 4- Horizontal velocity 5- Temperature 6- Surface loading
rate = 24 - 48 m³/m²/day
Primary sedimentation tank
(horizontal flow)
Effluent weir of rectangular sedimentation tank
Rectangular primary sedimentation tank
Primary sedimentation tank
(Radial flow)
Circular primary sedimentation tank
7- Dimension of tank 8- Dead zones.
Design criteria: 1- Retention period = T = 2 - 3 hrs 2- Surface
loading rate (S.L.R.) = 24 - 48 m³/m²/day 3- Horizontal velocity ≤
0.3 m/min 4- Effluent weir loading (E.W.L.) ≤ 600 m³/m/day (≤ 25
m³/m/hr) 5- L = 3 – 5 B
L ≤ 40 m 6- d = 3 - 5 m 7- B = 2 – 3 d
8- Φ ≤ 40 m9- Bottom slope for circular tank = 4 - 10 %
for rectangular tank = 1 - 2 %
V = Qd x T
S.L.R = Qd / S.A
Example:
For a sewage treatment plant, the following data are given:
- Qave summer = 18000 m3/d
- S.L.R = 30 m3/m2/d
It is required to design primary sedimentation tanks.
Solution:
Qd = 1.5 x 18000 = 27000 m3/d
For rectangular tank:
Assume T = 2.5 hr
≤L/7
≤L/7
Circular primary sedimentation tank:
Example:
Estimate the volume of sludge produced per 27000 m3/d if the
influent S.S 300 mg/l. The removal efficiency is 60%.
Solution:
For circular tank:
89
2
3
900
30
27000
.
.
.
.
5
.
2812
24
5
.
2
27000
m
A
S
A
S
Q
R
L
S
m
T
Q
V
d
d
=
=
=
=
´
=
´
=
safe
d
m
m
n
n
Q
L
W
E
safe
m
m
d
n
Q
v
Check
d
d
h
600
/
/
179
24
27000
.
.
2
min
/
3
.
0
min
/
04
.
0
)
60
24
(
1
.
3
24
2
27000
1
:
3
p
p
=
´
´
=
´
´
=
-
=
´
´
´
´
´
=
´
´
´
=
-
p
f
p
p
f
p
m
loading
weir
Q
weir
of
Length
d
m
m
loading
weir
take
unsafe
d
m
m
b
n
Q
L
W
E
safe
m
d
b
n
Q
area
tional
cross
Q
V
Check
m
L
A
S
b
m
length
Assume
m
k
one
of
A
S
n
Take
m
A
S
V
d
d
d
d
d
h
5
.
22
600
2
27000
/
/
600
600
/
/
1200
25
.
11
2
27000
.
.
2
min
/
3
.
0
269
.
0
60
24
)
1
.
3
25
.
11
2
(
27000
sec
1
:
25
.
11
40
450
.
40
450
2
900
tan
.
2
1
.
3
900
5
.
2812
.
3
3
2
=
´
=
=
=
\
=
´
=
´
=
-
=
´
´
´
´
=
´
´
=
=
-
=
=
=
=
=
=
\
=
=
=
=
f
p
2
3
900
30
2700
.
.
.
.
5
.
2812
24
5
.
2
27000
m
A
S
A
S
Q
R
L
S
m
T
Q
V
d
d
=
=
=
=
´
=
´
=
m
m
k
one
of
A
S
n
Take
m
A
S
V
d
24
4
450
2
900
tan
.
2
1
.
3
900
2700
.
2
2
=
\
=
=
=
\
=
=
=
=
f
pf
d
m
G
S
weight
solids
dry
of
Volume
m
t
sludge
of
gravity
Specific
d
t
solids
dry
of
Weight
/
718
.
4
03
.
1
86
.
4
.
/
03
.
1
/
86
.
4
10
27000
300
100
60
3
3
6
=
=
=
=
=
´
´
´
=
-
)
2
1
(
925
.
0
265
.
38
3
796
.
11
)
64
.
31
1
64
.
31
1
(
3
796
.
11
64
.
31
2
25
.
11
2
25
.
11
2
2
1
1
)
(
3
/
796
.
11
2
59
.
23
12
min
/
59
.
23
2
18
.
47
/
18
.
47
2
37
.
94
tan
:
/
37
.
94
05
.
0
718
.
4
)
%
5
(
%
95
2
2
1
2
1
2
1
3
3
3
3
-
=
=
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=
+
+
+
=
=
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=
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=
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=
+
+
+
=
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=
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=
h
m
h
h
m
B
B
a
m
m
a
a
a
a
a
h
V
d
m
hopper
per
sludge
of
Volume
hours
every
withdrawal
sludge
g
Assu
d
m
hopper
per
sludge
of
Volume
d
m
k
per
sludge
of
Volume
hopper
of
Design
d
m
sludge
of
Volume
solids
sludge
of
content
Moisture
o
o
q
f
f
f
p
60
45
2
1
1
)
2
(
4
2
2
-
=
-
=
³
+
=
m
h
m
h
V
s
s
b
safe
s
m
v
mm
mm
m
s
m
pipe
sludge
in
v
v
A
Q
s
m
Q
utes
utes
withdrawal
of
time
Assume
pipe
withdrawal
sludge
of
Design
act
sludge
sludge
/
24
.
1
)
150
(
200
22
.
0
1
4
039
.
0
/
5
.
1
1
/
039
.
0
60
5
796
.
11
)
min
20
5
(
min
5
:
2
3
=
³
»
=
\
´
=
-
=
´
=
=
´
=
-
f
f
pf
